Find the Maximum/Minimum Value g(t)=t square root of t+3

$g (t) = t \sqrt{t + 3}$

Step 1

Find the first derivative of the function.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t + 3}$ as ${(t + 3)}^{\frac{1}{2}}$ .

$\frac{d}{d t} [t {(t + 3)}^{\frac{1}{2}}]$

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = {(t + 3)}^{\frac{1}{2}}$ .

$t \frac{d}{d t} [{(t + 3)}^{\frac{1}{2}}] + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{\frac{1}{2}}$ and $g (t) = t + 3$ .

To apply the Chain Rule, set $u$ as $t + 3$ .

$t (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$t (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Replace all occurrences of $u$ with $t + 3$ .

$t (\frac{1}{2} {(t + 3)}^{\frac{1}{2} - 1} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$t (\frac{1}{2} {(t + 3)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Combine $- 1$ and $\frac{2}{2}$ .

$t (\frac{1}{2} {(t + 3)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Combine the numerators over the common denominator.

$t (\frac{1}{2} {(t + 3)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$t (\frac{1}{2} {(t + 3)}^{\frac{1 - 2}{2}} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Subtract $2$ from $1$ .

$t (\frac{1}{2} {(t + 3)}^{\frac{- 1}{2}} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Combine fractions.

Move the negative in front of the fraction.

$t (\frac{1}{2} {(t + 3)}^{- \frac{1}{2}} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Combine $\frac{1}{2}$ and ${(t + 3)}^{- \frac{1}{2}}$ .

$t (\frac{{(t + 3)}^{- \frac{1}{2}}}{2} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Move ${(t + 3)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$t (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}} \frac{d}{d t} [t + 3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Combine $\frac{1}{2 {(t + 3)}^{\frac{1}{2}}}$ and $t$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} \frac{d}{d t} [t + 3] + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

By the Sum Rule, the derivative of $t + 3$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [3]$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} (\frac{d}{d t} [t] + \frac{d}{d t} [3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} (1 + \frac{d}{d t} [3]) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} (1 + 0) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Simplify the expression.

Add $1$ and $0$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} \cdot 1 + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Multiply $\frac{t}{2 {(t + 3)}^{\frac{1}{2}}}$ by $1$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} [t]$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + {(t + 3)}^{\frac{1}{2}} \cdot 1$

Multiply ${(t + 3)}^{\frac{1}{2}}$ by $1$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + {(t + 3)}^{\frac{1}{2}}$

To write ${(t + 3)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{2 {(t + 3)}^{\frac{1}{2}}}{2 {(t + 3)}^{\frac{1}{2}}}$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + {(t + 3)}^{\frac{1}{2}} \cdot \frac{2 {(t + 3)}^{\frac{1}{2}}}{2 {(t + 3)}^{\frac{1}{2}}}$

Combine ${(t + 3)}^{\frac{1}{2}}$ and $\frac{2 {(t + 3)}^{\frac{1}{2}}}{2 {(t + 3)}^{\frac{1}{2}}}$ .

$\frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + \frac{{(t + 3)}^{\frac{1}{2}} (2 {(t + 3)}^{\frac{1}{2}})}{2 {(t + 3)}^{\frac{1}{2}}}$

Combine the numerators over the common denominator.

$\frac{t + {(t + 3)}^{\frac{1}{2}} (2 {(t + 3)}^{\frac{1}{2}})}{2 {(t + 3)}^{\frac{1}{2}}}$

Multiply ${(t + 3)}^{\frac{1}{2}}$ by ${(t + 3)}^{\frac{1}{2}}$ by adding the exponents.

Move ${(t + 3)}^{\frac{1}{2}}$ .

$\frac{t + {(t + 3)}^{\frac{1}{2}} {(t + 3)}^{\frac{1}{2}} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{t + {(t + 3)}^{\frac{1}{2} + \frac{1}{2}} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Combine the numerators over the common denominator.

$\frac{t + {(t + 3)}^{\frac{1 + 1}{2}} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Add $1$ and $1$ .

$\frac{t + {(t + 3)}^{\frac{2}{2}} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Divide $2$ by $2$ .

$\frac{t + {(t + 3)}^{1} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Simplify ${(t + 3)}^{1} \cdot 2$ .

$\frac{t + (t + 3) \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Move $2$ to the left of $t + 3$ .

$\frac{t + 2 \cdot (t + 3)}{2 {(t + 3)}^{\frac{1}{2}}}$

Simplify.

Apply the distributive property.

$\frac{t + 2 t + 2 \cdot 3}{2 {(t + 3)}^{\frac{1}{2}}}$

Simplify the numerator.

Multiply $2$ by $3$ .

$\frac{t + 2 t + 6}{2 {(t + 3)}^{\frac{1}{2}}}$

Add $t$ and $2 t$ .

$\frac{3 t + 6}{2 {(t + 3)}^{\frac{1}{2}}}$

Factor $3$ out of $3 t + 6$ .

Factor $3$ out of $3 t$ .

$\frac{3 (t) + 6}{2 {(t + 3)}^{\frac{1}{2}}}$

Factor $3$ out of $6$ .

$\frac{3 t + 3 \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Factor $3$ out of $3 t + 3 \cdot 2$ .

$\frac{3 (t + 2)}{2 {(t + 3)}^{\frac{1}{2}}}$

Step 2

Find the second derivative of the function.

Since $\frac{3}{2}$ is constant with respect to $t$ , the derivative of $\frac{3 (t + 2)}{2 {(t + 3)}^{\frac{1}{2}}}$ with respect to $t$ is $\frac{3}{2} \frac{d}{d t} [\frac{t + 2}{{(t + 3)}^{\frac{1}{2}}}]$ .

$g'' (t) = \frac{3}{2} \cdot \frac{d}{d t} (\frac{t + 2}{{(t + 3)}^{\frac{1}{2}}})$

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = t + 2$ and $g (t) = {(t + 3)}^{\frac{1}{2}}$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t + 2) - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{{({(t + 3)}^{\frac{1}{2}})}^{2}}$

Multiply the exponents in ${({(t + 3)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t + 2) - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{{(t + 3)}^{\frac{1}{2} \cdot 2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t + 2) - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{{(t + 3)}^{\frac{1}{2} \cdot 2}}$

Rewrite the expression.

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t + 2) - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{t + 3}$

Simplify.

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t + 2) - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{t + 3}$

Differentiate.

By the Sum Rule, the derivative of $t + 2$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [2]$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} (\frac{d}{d t} (t) + \frac{d}{d t} (2)) - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{t + 3}$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} (1 + \frac{d}{d t} (2)) - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{t + 3}$

Since $2$ is constant with respect to $t$ , the derivative of $2$ with respect to $t$ is $0$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} (1 + 0) - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{t + 3}$

Simplify the expression.

Add $1$ and $0$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} \cdot 1 - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{t + 3}$

Multiply ${(t + 3)}^{\frac{1}{2}}$ by $1$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) \frac{d}{d t} {(t + 3)}^{\frac{1}{2}}}{t + 3}$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{\frac{1}{2}}$ and $g (t) = t + 3$ .

To apply the Chain Rule, set $u_{1}$ as $t + 3$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{d}{d u (1)} ({u_{1}}^{\frac{1}{2}}) \frac{d}{d t} (t + 3))}{t + 3}$

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = \frac{1}{2}$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2} {u_{1}}^{\frac{1}{2} - 1} \frac{d}{d t} (t + 3))}{t + 3}$

Replace all occurrences of $u_{1}$ with $t + 3$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2} \cdot {(t + 3)}^{\frac{1}{2} - 1} \frac{d}{d t} (t + 3))}{t + 3}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2} \cdot {(t + 3)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d t} (t + 3))}{t + 3}$

Combine $- 1$ and $\frac{2}{2}$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2} \cdot {(t + 3)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d t} (t + 3))}{t + 3}$

Combine the numerators over the common denominator.

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2} \cdot {(t + 3)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d t} (t + 3))}{t + 3}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2} \cdot {(t + 3)}^{\frac{1 - 2}{2}} \frac{d}{d t} (t + 3))}{t + 3}$

Subtract $2$ from $1$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2} \cdot {(t + 3)}^{\frac{- 1}{2}} \frac{d}{d t} (t + 3))}{t + 3}$

Combine fractions.

Move the negative in front of the fraction.

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2} \cdot {(t + 3)}^{- \frac{1}{2}} \frac{d}{d t} (t + 3))}{t + 3}$

Combine $\frac{1}{2}$ and ${(t + 3)}^{- \frac{1}{2}}$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{{(t + 3)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d t} (t + 3))}{t + 3}$

Move ${(t + 3)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}} \cdot \frac{d}{d t} (t + 3))}{t + 3}$

By the Sum Rule, the derivative of $t + 3$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [3]$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}} \cdot (\frac{d}{d t} (t) + \frac{d}{d t} (3)))}{t + 3}$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}} \cdot (1 + \frac{d}{d t} (3)))}{t + 3}$

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}} \cdot (1 + 0))}{t + 3}$

Combine fractions.

Add $1$ and $0$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}} \cdot 1)}{t + 3}$

Multiply $\frac{1}{2 {(t + 3)}^{\frac{1}{2}}}$ by $1$ .

$g'' (t) = \frac{3}{2} \cdot \frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) \frac{1}{2 {(t + 3)}^{\frac{1}{2}}}}{t + 3}$

Multiply $\frac{3}{2}$ by $\frac{{(t + 3)}^{\frac{1}{2}} - (t + 2) \frac{1}{2 {(t + 3)}^{\frac{1}{2}}}}{t + 3}$ .

$g'' (t) = \frac{3 ({(t + 3)}^{\frac{1}{2}} - (t + 2) \frac{1}{2 {(t + 3)}^{\frac{1}{2}}})}{2 (t + 3)}$

Simplify.

Apply the distributive property.

$g'' (t) = \frac{3 ({(t + 3)}^{\frac{1}{2}} + (- t - 1 \cdot 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}}))}{2 (t + 3)}$

Apply the distributive property.

$g'' (t) = \frac{3 {(t + 3)}^{\frac{1}{2}} + 3 ((- t - 1 \cdot 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}}))}{2 (t + 3)}$

Apply the distributive property.

$g'' (t) = \frac{3 {(t + 3)}^{\frac{1}{2}} + 3 ((- t - 1 \cdot 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}}))}{2 t + 2 \cdot 3}$

Simplify the numerator.

Factor $3$ out of $3 {(t + 3)}^{\frac{1}{2}} + 3 (- t - 1 \cdot 2) \frac{1}{2 {(t + 3)}^{\frac{1}{2}}}$ .

Factor $3$ out of $3 {(t + 3)}^{\frac{1}{2}}$ .

$g'' (t) = \frac{3 ({(t + 3)}^{\frac{1}{2}}) + 3 (- t - 1 \cdot 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Factor $3$ out of $3 (- t - 1 \cdot 2) \frac{1}{2 {(t + 3)}^{\frac{1}{2}}}$ .

$g'' (t) = \frac{3 ({(t + 3)}^{\frac{1}{2}}) + 3 ((- t - 1 \cdot 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}}))}{2 t + 2 \cdot 3}$

Factor $3$ out of $3 ({(t + 3)}^{\frac{1}{2}}) + 3 ((- t - 1 \cdot 2) \frac{1}{2 {(t + 3)}^{\frac{1}{2}}})$ .

$g'' (t) = \frac{3 ({(t + 3)}^{\frac{1}{2}} + (- t - 1 \cdot 2) (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}}))}{2 t + 2 \cdot 3}$

Let $u_{2} = {(t + 3)}^{\frac{1}{2}}$ . Substitute $u_{2}$ for all occurrences of ${(t + 3)}^{\frac{1}{2}}$ .

Rewrite using the commutative property of multiplication.

$g'' (t) = \frac{3 (\frac{2 u_{2} u_{2} - t - 2}{2 u_{2}})}{2 t + 2 \cdot 3}$

Multiply $u_{2}$ by $u_{2}$ by adding the exponents.

Move $u_{2}$ .

$g'' (t) = \frac{3 (\frac{2 (u_{2} u_{2}) - t - 2}{2 u_{2}})}{2 t + 2 \cdot 3}$

Multiply $u_{2}$ by $u_{2}$ .

$g'' (t) = \frac{3 (\frac{2 {u_{2}}^{2} - t - 2}{2 u_{2}})}{2 t + 2 \cdot 3}$

Replace all occurrences of $u_{2}$ with ${(t + 3)}^{\frac{1}{2}}$ .

$g'' (t) = \frac{3 (\frac{2 {({(t + 3)}^{\frac{1}{2}})}^{2} - t - 2}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Simplify.

Simplify each term.

Multiply the exponents in ${({(t + 3)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$g'' (t) = \frac{3 (\frac{2 {(t + 3)}^{\frac{1}{2} \cdot 2} - t - 2}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Cancel the common factor of $2$ .

Cancel the common factor.

$g'' (t) = \frac{3 (\frac{2 {(t + 3)}^{\frac{1}{2} \cdot 2} - t - 2}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Rewrite the expression.

$g'' (t) = \frac{3 (\frac{2 (t + 3) - t - 2}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Simplify.

$g'' (t) = \frac{3 (\frac{2 (t + 3) - t - 2}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Apply the distributive property.

$g'' (t) = \frac{3 (\frac{2 t + 2 \cdot 3 - t - 2}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Multiply $2$ by $3$ .

$g'' (t) = \frac{3 (\frac{2 t + 6 - t - 2}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Subtract $t$ from $2 t$ .

$g'' (t) = \frac{3 (\frac{t + 6 - 2}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Subtract $2$ from $6$ .

$g'' (t) = \frac{3 (\frac{t + 4}{2 {(t + 3)}^{\frac{1}{2}}})}{2 t + 2 \cdot 3}$

Combine terms.

Combine $3$ and $\frac{t + 4}{2 {(t + 3)}^{\frac{1}{2}}}$ .

$g'' (t) = \frac{\frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}}}}{2 t + 2 \cdot 3}$

Multiply $2$ by $3$ .

$g'' (t) = \frac{\frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}}}}{2 t + 6}$

Rewrite $\frac{\frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}}}}{2 t + 6}$ as a product.

$g'' (t) = \frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}}} \cdot \frac{1}{2 t + 6}$

Multiply $\frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}}}$ by $\frac{1}{2 t + 6}$ .

$g'' (t) = \frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}} (2 t + 6)}$

Simplify the denominator.

Factor $2$ out of $2 t + 6$ .

Factor $2$ out of $2 t$ .

$g'' (t) = \frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}} (2 (t) + 6)}$

Factor $2$ out of $6$ .

$g'' (t) = \frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}} (2 t + 2 \cdot 3)}$

Factor $2$ out of $2 t + 2 \cdot 3$ .

$g'' (t) = \frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}} (2 (t + 3))}$

$g'' (t) = \frac{3 (t + 4)}{2 {(t + 3)}^{\frac{1}{2}} \cdot (2 (t + 3))}$

Combine exponents.

Multiply $2$ by $2$ .

$g'' (t) = \frac{3 (t + 4)}{4 {(t + 3)}^{\frac{1}{2}} (t + 3)}$

Raise $t + 3$ to the power of $1$ .

$g'' (t) = \frac{3 (t + 4)}{4 ((t + 3) {(t + 3)}^{\frac{1}{2}})}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$g'' (t) = \frac{3 (t + 4)}{4 {(t + 3)}^{1 + \frac{1}{2}}}$

Write $1$ as a fraction with a common denominator.

$g'' (t) = \frac{3 (t + 4)}{4 {(t + 3)}^{\frac{2}{2} + \frac{1}{2}}}$

Combine the numerators over the common denominator.

$g'' (t) = \frac{3 (t + 4)}{4 {(t + 3)}^{\frac{2 + 1}{2}}}$

Add $2$ and $1$ .

$g'' (t) = \frac{3 (t + 4)}{4 {(t + 3)}^{\frac{3}{2}}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\frac{3 (t + 2)}{2 {(t + 3)}^{\frac{1}{2}}} = 0$

Step 4

Find the first derivative.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t + 3}$ as ${(t + 3)}^{\frac{1}{2}}$ .

$f' (t) = \frac{d}{d t} (t {(t + 3)}^{\frac{1}{2}})$

$f' (t) = t \frac{d}{d t} ({(t + 3)}^{\frac{1}{2}}) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{\frac{1}{2}}$ and $g (t) = t + 3$ .

To apply the Chain Rule, set $u$ as $t + 3$ .

$f' (t) = t (\frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$f' (t) = t (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Replace all occurrences of $u$ with $t + 3$ .

$f' (t) = t (\frac{1}{2} \cdot {(t + 3)}^{\frac{1}{2} - 1} \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (t) = t (\frac{1}{2} \cdot {(t + 3)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine $- 1$ and $\frac{2}{2}$ .

$f' (t) = t (\frac{1}{2} \cdot {(t + 3)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine the numerators over the common denominator.

$f' (t) = t (\frac{1}{2} \cdot {(t + 3)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$f' (t) = t (\frac{1}{2} \cdot {(t + 3)}^{\frac{1 - 2}{2}} \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Subtract $2$ from $1$ .

$f' (t) = t (\frac{1}{2} \cdot {(t + 3)}^{\frac{- 1}{2}} \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine fractions.

Move the negative in front of the fraction.

$f' (t) = t (\frac{1}{2} \cdot {(t + 3)}^{- \frac{1}{2}} \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine $\frac{1}{2}$ and ${(t + 3)}^{- \frac{1}{2}}$ .

$f' (t) = t (\frac{{(t + 3)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Move ${(t + 3)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (t) = t (\frac{1}{2 {(t + 3)}^{\frac{1}{2}}} \cdot \frac{d}{d t} (t + 3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Combine $\frac{1}{2 {(t + 3)}^{\frac{1}{2}}}$ and $t$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} \cdot \frac{d}{d t} (t + 3) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

By the Sum Rule, the derivative of $t + 3$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [3]$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} \cdot (\frac{d}{d t} (t) + \frac{d}{d t} (3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} \cdot (1 + \frac{d}{d t} (3)) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} \cdot (1 + 0) + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Simplify the expression.

Add $1$ and $0$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} \cdot 1 + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Multiply $\frac{t}{2 {(t + 3)}^{\frac{1}{2}}}$ by $1$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + {(t + 3)}^{\frac{1}{2}} \frac{d}{d t} (t)$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + {(t + 3)}^{\frac{1}{2}} \cdot 1$

Multiply ${(t + 3)}^{\frac{1}{2}}$ by $1$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + {(t + 3)}^{\frac{1}{2}}$

To write ${(t + 3)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{2 {(t + 3)}^{\frac{1}{2}}}{2 {(t + 3)}^{\frac{1}{2}}}$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + {(t + 3)}^{\frac{1}{2}} \cdot \frac{2 {(t + 3)}^{\frac{1}{2}}}{2 {(t + 3)}^{\frac{1}{2}}}$

Combine ${(t + 3)}^{\frac{1}{2}}$ and $\frac{2 {(t + 3)}^{\frac{1}{2}}}{2 {(t + 3)}^{\frac{1}{2}}}$ .

$f' (t) = \frac{t}{2 {(t + 3)}^{\frac{1}{2}}} + \frac{{(t + 3)}^{\frac{1}{2}} (2 {(t + 3)}^{\frac{1}{2}})}{2 {(t + 3)}^{\frac{1}{2}}}$

Combine the numerators over the common denominator.

$f' (t) = \frac{t + {(t + 3)}^{\frac{1}{2}} (2 {(t + 3)}^{\frac{1}{2}})}{2 {(t + 3)}^{\frac{1}{2}}}$

Multiply ${(t + 3)}^{\frac{1}{2}}$ by ${(t + 3)}^{\frac{1}{2}}$ by adding the exponents.

Move ${(t + 3)}^{\frac{1}{2}}$ .

$f' (t) = \frac{t + {(t + 3)}^{\frac{1}{2}} {(t + 3)}^{\frac{1}{2}} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (t) = \frac{t + {(t + 3)}^{\frac{1}{2} + \frac{1}{2}} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Combine the numerators over the common denominator.

$f' (t) = \frac{t + {(t + 3)}^{\frac{1 + 1}{2}} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Add $1$ and $1$ .

$f' (t) = \frac{t + {(t + 3)}^{\frac{2}{2}} \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Divide $2$ by $2$ .

$f' (t) = \frac{t + (t + 3) \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Simplify ${(t + 3)}^{1} \cdot 2$ .

$f' (t) = \frac{t + (t + 3) \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Move $2$ to the left of $t + 3$ .

$f' (t) = \frac{t + 2 \cdot (t + 3)}{2 {(t + 3)}^{\frac{1}{2}}}$

Simplify.

Apply the distributive property.

$f' (t) = \frac{t + 2 t + 2 \cdot 3}{2 {(t + 3)}^{\frac{1}{2}}}$

Simplify the numerator.

Multiply $2$ by $3$ .

$f' (t) = \frac{t + 2 t + 6}{2 {(t + 3)}^{\frac{1}{2}}}$

Add $t$ and $2 t$ .

$f' (t) = \frac{3 t + 6}{2 {(t + 3)}^{\frac{1}{2}}}$

Factor $3$ out of $3 t + 6$ .

Factor $3$ out of $3 t$ .

$f' (t) = \frac{3 (t) + 6}{2 {(t + 3)}^{\frac{1}{2}}}$

Factor $3$ out of $6$ .

$f' (t) = \frac{3 t + 3 \cdot 2}{2 {(t + 3)}^{\frac{1}{2}}}$

Factor $3$ out of $3 t + 3 \cdot 2$ .

$f' (t) = \frac{3 (t + 2)}{2 {(t + 3)}^{\frac{1}{2}}}$

The first derivative of $g (t)$ with respect to $t$ is $\frac{3 (t + 2)}{2 {(t + 3)}^{\frac{1}{2}}}$ .

$\frac{3 (t + 2)}{2 {(t + 3)}^{\frac{1}{2}}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{3 (t + 2)}{2 {(t + 3)}^{\frac{1}{2}}} = 0$ .

Set the first derivative equal to $0$ .

$\frac{3 (t + 2)}{2 {(t + 3)}^{\frac{1}{2}}} = 0$

Set the numerator equal to zero.

$3 (t + 2) = 0$

Solve the equation for $t$ .

Divide each term in $3 (t + 2) = 0$ by $3$ and simplify.

Divide each term in $3 (t + 2) = 0$ by $3$ .

$\frac{3 (t + 2)}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (t + 2)}{3} = \frac{0}{3}$

Divide $t + 2$ by $1$ .

$t + 2 = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$t + 2 = 0$

Subtract $2$ from both sides of the equation.

$t = - 2$

Step 6

Find the values where the derivative is undefined.

Convert expressions with fractional exponents to radicals.

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{3 (t + 2)}{2 \sqrt{{(t + 3)}^{1}}}$

Anything raised to $1$ is the base itself.

$\frac{3 (t + 2)}{2 \sqrt{t + 3}}$

Set the denominator in $\frac{3 (t + 2)}{2 \sqrt{t + 3}}$ equal to $0$ to find where the expression is undefined.

$2 \sqrt{t + 3} = 0$

Solve for $t$ .

To remove the radical on the left side of the equation, square both sides of the equation.

${(2 \sqrt{t + 3})}^{2} = 0^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t + 3}$ as ${(t + 3)}^{\frac{1}{2}}$ .

${(2 {(t + 3)}^{\frac{1}{2}})}^{2} = 0^{2}$

Simplify the left side.

Simplify ${(2 {(t + 3)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $2 {(t + 3)}^{\frac{1}{2}}$ .

$2^{2} {({(t + 3)}^{\frac{1}{2}})}^{2} = 0^{2}$

Raise $2$ to the power of $2$ .

$4 {({(t + 3)}^{\frac{1}{2}})}^{2} = 0^{2}$

Multiply the exponents in ${({(t + 3)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 {(t + 3)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(t + 3)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Rewrite the expression.

$4 {(t + 3)}^{1} = 0^{2}$

Simplify.

$4 (t + 3) = 0^{2}$

Apply the distributive property.

$4 t + 4 \cdot 3 = 0^{2}$

Multiply $4$ by $3$ .

$4 t + 12 = 0^{2}$

Simplify the right side.

Raising $0$ to any positive power yields $0$ .

$4 t + 12 = 0$

Solve for $t$ .

Subtract $12$ from both sides of the equation.

$4 t = - 12$

Divide each term in $4 t = - 12$ by $4$ and simplify.

Divide each term in $4 t = - 12$ by $4$ .

$\frac{4 t}{4} = \frac{- 12}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 t}{4} = \frac{- 12}{4}$

Divide $t$ by $1$ .

$t = \frac{- 12}{4}$

Simplify the right side.

Divide $- 12$ by $4$ .

$t = - 3$

Set the radicand in $\sqrt{t + 3}$ less than $0$ to find where the expression is undefined.

$t + 3 < 0$

Subtract $3$ from both sides of the inequality.

$t < - 3$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$t \leq - 3$

$(- \infty, - 3]$

$t \leq - 3$

$(- \infty, - 3]$

Step 7

Critical points to evaluate.

$t = - 2, - 3$

Step 8

Evaluate the second derivative at $t = - 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{3 ((- 2) + 4)}{4 {((- 2) + 3)}^{\frac{3}{2}}}$

Step 9

Evaluate the second derivative.

Factor $2$ out of $3 ((- 2) + 4)$ .

$\frac{2 (3 (- 1 + 2))}{4 {((- 2) + 3)}^{\frac{3}{2}}}$

Cancel the common factors.

Factor $2$ out of $4 {((- 2) + 3)}^{\frac{3}{2}}$ .

$\frac{2 (3 (- 1 + 2))}{2 (2 {((- 2) + 3)}^{\frac{3}{2}})}$

Cancel the common factor.

$\frac{2 (3 (- 1 + 2))}{2 (2 {((- 2) + 3)}^{\frac{3}{2}})}$

Rewrite the expression.

$\frac{3 (- 1 + 2)}{2 {((- 2) + 3)}^{\frac{3}{2}}}$

Add $- 1$ and $2$ .

$\frac{3 \cdot 1}{2 {(- 2 + 3)}^{\frac{3}{2}}}$

Simplify the denominator.

Add $- 2$ and $3$ .

$\frac{3 \cdot 1}{2 \cdot 1^{\frac{3}{2}}}$

One to any power is one.

$\frac{3 \cdot 1}{2 \cdot 1}$

Simplify.

Multiply $3$ by $1$ .

$\frac{3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$\frac{3}{2}$

Step 10

$t = - 2$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$t = - 2$ is a local minimum

Step 11

Find the y-value when $t = - 2$ .

Replace the variable $t$ with $- 2$ in the expression.

$g (- 2) = (- 2) \sqrt{(- 2) + 3}$

Simplify the result.

Add $- 2$ and $3$ .

$g (- 2) = - 2 \sqrt{1}$

Any root of $1$ is $1$ .

$g (- 2) = - 2 \cdot 1$

Multiply $- 2$ by $1$ .

$g (- 2) = - 2$

The final answer is $- 2$ .

$y = - 2$

Step 12

Evaluate the second derivative at $t = - 3$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{3 ((- 3) + 4)}{4 {((- 3) + 3)}^{\frac{3}{2}}}$

Step 13

Evaluate the second derivative.

Simplify the expression.

Add $- 3$ and $3$ .

$\frac{3 ((- 3) + 4)}{4 \cdot 0^{\frac{3}{2}}}$

Rewrite $0$ as $0^{2}$ .

$\frac{3 ((- 3) + 4)}{4 \cdot {(0^{2})}^{\frac{3}{2}}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{3 ((- 3) + 4)}{4 \cdot 0^{2 (\frac{3}{2})}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{3 ((- 3) + 4)}{4 \cdot 0^{2 (\frac{3}{2})}}$

Rewrite the expression.

$\frac{3 ((- 3) + 4)}{4 \cdot 0^{3}}$

Simplify the expression.

Raising $0$ to any positive power yields $0$ .

$\frac{3 ((- 3) + 4)}{4 \cdot 0}$

Multiply $4$ by $0$ .

$\frac{3 ((- 3) + 4)}{0}$

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{3 ((- 3) + 4)}{0}$

The expression contains a division by $0$ . The expression is undefined.

Undefined

Step 14

Since the first derivative test failed, there are no local extrema.

No Local Extrema

Step 15

Find the Maximum/Minimum Value g(t)=t square root of t+3

Related Questions

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