Find the Maximum/Minimum Value y=3|x+6|+2

$y = 3 | x + 6 | + 2$

Step 1

Find the first derivative of the function.

By the Sum Rule, the derivative of $3 | x + 6 | + 2$ with respect to $x$ is $\frac{d}{d x} [3 | x + 6 |] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [3 | x + 6 |] + \frac{d}{d x} [2]$

Evaluate $\frac{d}{d x} [3 | x + 6 |]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 | x + 6 |$ with respect to $x$ is $3 \frac{d}{d x} [| x + 6 |]$ .

$3 \frac{d}{d x} [| x + 6 |] + \frac{d}{d x} [2]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = x + 6$ .

To apply the Chain Rule, set $u$ as $x + 6$ .

$3 (\frac{d}{d u} [| u |] \frac{d}{d x} [x + 6]) + \frac{d}{d x} [2]$

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$3 (\frac{u}{| u |} \frac{d}{d x} [x + 6]) + \frac{d}{d x} [2]$

Replace all occurrences of $u$ with $x + 6$ .

$3 (\frac{x + 6}{| x + 6 |} \frac{d}{d x} [x + 6]) + \frac{d}{d x} [2]$

By the Sum Rule, the derivative of $x + 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [6]$ .

$3 (\frac{x + 6}{| x + 6 |} (\frac{d}{d x} [x] + \frac{d}{d x} [6])) + \frac{d}{d x} [2]$

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

$3 (\frac{x + 6}{| x + 6 |} (1 + \frac{d}{d x} [6])) + \frac{d}{d x} [2]$

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$3 (\frac{x + 6}{| x + 6 |} (1 + 0)) + \frac{d}{d x} [2]$

Add $1$ and $0$ .

$3 (\frac{x + 6}{| x + 6 |} \cdot 1) + \frac{d}{d x} [2]$

Multiply $\frac{x + 6}{| x + 6 |}$ by $1$ .

$3 \frac{x + 6}{| x + 6 |} + \frac{d}{d x} [2]$

Combine $3$ and $\frac{x + 6}{| x + 6 |}$ .

$\frac{3 (x + 6)}{| x + 6 |} + \frac{d}{d x} [2]$

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

$\frac{3 (x + 6)}{| x + 6 |} + 0$

Simplify.

Apply the distributive property.

$\frac{3 x + 3 \cdot 6}{| x + 6 |} + 0$

Combine terms.

Multiply $3$ by $6$ .

$\frac{3 x + 18}{| x + 6 |} + 0$

Add $\frac{3 x + 18}{| x + 6 |}$ and $0$ .

$\frac{3 x + 18}{| x + 6 |}$

Factor $3$ out of $3 x + 18$ .

Factor $3$ out of $3 x$ .

$\frac{3 (x) + 18}{| x + 6 |}$

Factor $3$ out of $18$ .

$\frac{3 x + 3 \cdot 6}{| x + 6 |}$

Factor $3$ out of $3 x + 3 \cdot 6$ .

$\frac{3 (x + 6)}{| x + 6 |}$

Step 2

Find the second derivative of the function.

Since $3$ is constant with respect to $x$ , the derivative of $\frac{3 (x + 6)}{| x + 6 |}$ with respect to $x$ is $3 \frac{d}{d x} [\frac{x + 6}{| x + 6 |}]$ .

$f'' (x) = 3 \frac{d}{d x} (\frac{x + 6}{| x + 6 |})$

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x + 6$ and $g (x) = | x + 6 |$ .

$f'' (x) = 3 (\frac{| x + 6 | \frac{d}{d x} (x + 6) - (x + 6) \frac{d}{d x} | x + 6 |}{{| x + 6 |}^{2}})$

Differentiate.

By the Sum Rule, the derivative of $x + 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [6]$ .

$f'' (x) = 3 (\frac{| x + 6 | (\frac{d}{d x} (x) + \frac{d}{d x} (6)) - (x + 6) \frac{d}{d x} | x + 6 |}{{| x + 6 |}^{2}})$

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

$f'' (x) = 3 (\frac{| x + 6 | (1 + \frac{d}{d x} (6)) - (x + 6) \frac{d}{d x} | x + 6 |}{{| x + 6 |}^{2}})$

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$f'' (x) = 3 (\frac{| x + 6 | (1 + 0) - (x + 6) \frac{d}{d x} | x + 6 |}{{| x + 6 |}^{2}})$

Simplify the expression.

Add $1$ and $0$ .

$f'' (x) = 3 (\frac{| x + 6 | \cdot 1 - (x + 6) \frac{d}{d x} | x + 6 |}{{| x + 6 |}^{2}})$

Multiply $| x + 6 |$ by $1$ .

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) \frac{d}{d x} | x + 6 |}{{| x + 6 |}^{2}})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = x + 6$ .

To apply the Chain Rule, set $u$ as $x + 6$ .

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) (\frac{d}{d u} (| u |) \frac{d}{d x} (x + 6))}{{| x + 6 |}^{2}})$

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) (\frac{u}{| u |} \cdot \frac{d}{d x} (x + 6))}{{| x + 6 |}^{2}})$

Replace all occurrences of $u$ with $x + 6$ .

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) (\frac{x + 6}{| x + 6 |} \cdot \frac{d}{d x} (x + 6))}{{| x + 6 |}^{2}})$

Differentiate.

By the Sum Rule, the derivative of $x + 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [6]$ .

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) (\frac{x + 6}{| x + 6 |} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (6)))}{{| x + 6 |}^{2}})$

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

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) (\frac{x + 6}{| x + 6 |} \cdot (1 + \frac{d}{d x} (6)))}{{| x + 6 |}^{2}})$

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) (\frac{x + 6}{| x + 6 |} \cdot (1 + 0))}{{| x + 6 |}^{2}})$

Combine fractions.

Add $1$ and $0$ .

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) (\frac{x + 6}{| x + 6 |} \cdot 1)}{{| x + 6 |}^{2}})$

Multiply $\frac{x + 6}{| x + 6 |}$ by $1$ .

$f'' (x) = 3 (\frac{| x + 6 | - (x + 6) \frac{x + 6}{| x + 6 |}}{{| x + 6 |}^{2}})$

Combine $3$ and $\frac{| x + 6 | - (x + 6) \frac{x + 6}{| x + 6 |}}{{| x + 6 |}^{2}}$ .

$f'' (x) = \frac{3 (| x + 6 | - (x + 6) \frac{x + 6}{| x + 6 |})}{{| x + 6 |}^{2}}$

Simplify.

Apply the distributive property.

$f'' (x) = \frac{3 (| x + 6 | + (- x - 1 \cdot 6) (\frac{x + 6}{| x + 6 |}))}{{| x + 6 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{3 | x + 6 | + 3 ((- x - 1 \cdot 6) (\frac{x + 6}{| x + 6 |}))}{{| x + 6 |}^{2}}$

Simplify the numerator.

Simplify each term.

Multiply $- 1$ by $6$ .

$f'' (x) = \frac{3 | x + 6 | + 3 ((- x - 6) (\frac{x + 6}{| x + 6 |}))}{{| x + 6 |}^{2}}$

Multiply $- x - 6$ by $\frac{x + 6}{| x + 6 |}$ .

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{(- x - 6) (x + 6)}{| x + 6 |})}{{| x + 6 |}^{2}}$

Simplify the numerator.

Factor $- 1$ out of $- x$ .

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{(- (x) - 6) (x + 6)}{| x + 6 |})}{{| x + 6 |}^{2}}$

Rewrite $- 6$ as $- 1 (6)$ .

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{(- (x) - 1 \cdot 6) (x + 6)}{| x + 6 |})}{{| x + 6 |}^{2}}$

Factor $- 1$ out of $- (x) - 1 (6)$ .

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{- (x + 6) (x + 6)}{| x + 6 |})}{{| x + 6 |}^{2}}$

Rewrite $- (x + 6)$ as $- 1 (x + 6)$ .

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{- 1 (x + 6) (x + 6)}{| x + 6 |})}{{| x + 6 |}^{2}}$

Raise $x + 6$ to the power of $1$ .

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{- 1 ((x + 6) (x + 6))}{| x + 6 |})}{{| x + 6 |}^{2}}$

Raise $x + 6$ to the power of $1$ .

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{- 1 ((x + 6) (x + 6))}{| x + 6 |})}{{| x + 6 |}^{2}}$

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

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{- 1 {(x + 6)}^{1 + 1}}{| x + 6 |})}{{| x + 6 |}^{2}}$

Add $1$ and $1$ .

$f'' (x) = \frac{3 | x + 6 | + 3 (\frac{- 1 {(x + 6)}^{2}}{| x + 6 |})}{{| x + 6 |}^{2}}$

Move the negative in front of the fraction.

$f'' (x) = \frac{3 | x + 6 | + 3 (- \frac{{(x + 6)}^{2}}{| x + 6 |})}{{| x + 6 |}^{2}}$

Multiply $3 (- \frac{{(x + 6)}^{2}}{| x + 6 |})$ .

Multiply $- 1$ by $3$ .

$f'' (x) = \frac{3 | x + 6 | - 3 \frac{{(x + 6)}^{2}}{| x + 6 |}}{{| x + 6 |}^{2}}$

Combine $- 3$ and $\frac{{(x + 6)}^{2}}{| x + 6 |}$ .

$f'' (x) = \frac{3 | x + 6 | + \frac{- 3 {(x + 6)}^{2}}{| x + 6 |}}{{| x + 6 |}^{2}}$

Move the negative in front of the fraction.

$f'' (x) = \frac{3 | x + 6 | - \frac{3 {(x + 6)}^{2}}{| x + 6 |}}{{| x + 6 |}^{2}}$

To write $3 | x + 6 |$ as a fraction with a common denominator, multiply by $\frac{| x + 6 |}{| x + 6 |}$ .

$f'' (x) = \frac{\frac{3 | x + 6 | | x + 6 |}{| x + 6 |} - \frac{3 {(x + 6)}^{2}}{| x + 6 |}}{{| x + 6 |}^{2}}$

Combine the numerators over the common denominator.

$f'' (x) = \frac{\frac{3 | x + 6 | | x + 6 | - 3 {(x + 6)}^{2}}{| x + 6 |}}{{| x + 6 |}^{2}}$

Simplify the numerator.

Factor $3$ out of $3 | x + 6 | | x + 6 | - 3 {(x + 6)}^{2}$ .

Factor $3$ out of $3 | x + 6 | | x + 6 |$ .

$f'' (x) = \frac{\frac{3 (| x + 6 | | x + 6 |) - 3 {(x + 6)}^{2}}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $3$ out of $- 3 {(x + 6)}^{2}$ .

$f'' (x) = \frac{\frac{3 (| x + 6 | | x + 6 |) + 3 (- {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $3$ out of $3 (| x + 6 | | x + 6 |) + 3 (- {(x + 6)}^{2})$ .

$f'' (x) = \frac{\frac{3 (| x + 6 | | x + 6 | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Multiply $| x + 6 | | x + 6 |$ .

To multiply absolute values, multiply the terms inside each absolute value.

$f'' (x) = \frac{\frac{3 (| (x + 6) (x + 6) | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Raise $x + 6$ to the power of $1$ .

$f'' (x) = \frac{\frac{3 (| (x + 6) (x + 6) | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Raise $x + 6$ to the power of $1$ .

$f'' (x) = \frac{\frac{3 (| (x + 6) (x + 6) | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

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

$f'' (x) = \frac{\frac{3 (| {(x + 6)}^{1 + 1} | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Add $1$ and $1$ .

$f'' (x) = \frac{\frac{3 (| {(x + 6)}^{2} | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Rewrite ${(x + 6)}^{2}$ as $(x + 6) (x + 6)$ .

$f'' (x) = \frac{\frac{3 (| (x + 6) (x + 6) | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Expand $(x + 6) (x + 6)$ using the FOIL Method.

Apply the distributive property.

$f'' (x) = \frac{\frac{3 (| x (x + 6) + 6 (x + 6) | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{\frac{3 (| x \cdot x + x \cdot 6 + 6 (x + 6) | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{\frac{3 (| x \cdot x + x \cdot 6 + 6 x + 6 \cdot 6 | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + x \cdot 6 + 6 x + 6 \cdot 6 | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Move $6$ to the left of $x$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 6 \cdot x + 6 x + 6 \cdot 6 | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Multiply $6$ by $6$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 6 x + 6 x + 36 | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Add $6 x$ and $6 x$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - {(x + 6)}^{2})}{| x + 6 |}}{{| x + 6 |}^{2}}$

Rewrite ${(x + 6)}^{2}$ as $(x + 6) (x + 6)$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - ((x + 6) (x + 6)))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Expand $(x + 6) (x + 6)$ using the FOIL Method.

Apply the distributive property.

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - (x (x + 6) + 6 (x + 6)))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - (x \cdot x + x \cdot 6 + 6 (x + 6)))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - (x \cdot x + x \cdot 6 + 6 x + 6 \cdot 6))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - (x^{2} + x \cdot 6 + 6 x + 6 \cdot 6))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Move $6$ to the left of $x$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - (x^{2} + 6 \cdot x + 6 x + 6 \cdot 6))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Multiply $6$ by $6$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - (x^{2} + 6 x + 6 x + 36))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Add $6 x$ and $6 x$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - (x^{2} + 12 x + 36))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - x^{2} - (12 x) - 1 \cdot 36)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Simplify.

Multiply $12$ by $- 1$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - x^{2} - 12 x - 1 \cdot 36)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Multiply $- 1$ by $36$ .

$f'' (x) = \frac{\frac{3 (| x^{2} + 12 x + 36 | - x^{2} - 12 x - 36)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Reorder terms.

$f'' (x) = \frac{\frac{3 (- x^{2} - 12 x + | x^{2} + 12 x + 36 | - 36)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Rewrite $- x^{2} - 12 x + | x^{2} + 12 x + 36 | - 36$ in a factored form.

Regroup terms.

$f'' (x) = \frac{\frac{3 (- 36 - x^{2} - 12 x + | x^{2} + 12 x + 36 |)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $- 1$ out of $- x^{2} - 12 x + | x^{2} + 12 x + 36 |$ .

Factor $- 1$ out of $- x^{2}$ .

$f'' (x) = \frac{\frac{3 (- 36 - (x^{2}) - 12 x + | x^{2} + 12 x + 36 |)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $- 1$ out of $- 12 x$ .

$f'' (x) = \frac{\frac{3 (- 36 - (x^{2}) - (12 x) + | x^{2} + 12 x + 36 |)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $- 1$ out of $| x^{2} + 12 x + 36 |$ .

$f'' (x) = \frac{\frac{3 (- 36 - (x^{2}) - (12 x) - 1 (- | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $- 1$ out of $- (x^{2}) - (12 x)$ .

$f'' (x) = \frac{\frac{3 (- 36 - (x^{2} + 12 x) - 1 (- | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $- 1$ out of $- (x^{2} + 12 x) - 1 (- | x^{2} + 12 x + 36 |)$ .

$f'' (x) = \frac{\frac{3 (- 36 - (x^{2} + 12 x - | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $- 1$ out of $- 36 - (x^{2} + 12 x - | x^{2} + 12 x + 36 |)$ .

Rewrite $- 36$ as $- 1 (36)$ .

$f'' (x) = \frac{\frac{3 (- 1 \cdot 36 - (x^{2} + 12 x - | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Factor $- 1$ out of $- 1 (36) - (x^{2} + 12 x - | x^{2} + 12 x + 36 |)$ .

$f'' (x) = \frac{\frac{3 (- 1 (36 + x^{2} + 12 x - | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Rewrite $- 1 (36 + x^{2} + 12 x - | x^{2} + 12 x + 36 |)$ as $- (36 + x^{2} + 12 x - | x^{2} + 12 x + 36 |)$ .

$f'' (x) = \frac{\frac{3 (- (36 + x^{2} + 12 x - | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Reorder terms.

$f'' (x) = \frac{\frac{3 (- (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

$f'' (x) = \frac{\frac{3 \cdot (- 1 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Combine exponents.

Factor out negative.

$f'' (x) = \frac{\frac{- (3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |))}{| x + 6 |}}{{| x + 6 |}^{2}}$

Multiply $3$ by $- 1$ .

$f'' (x) = \frac{\frac{- 3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Move the negative in front of the fraction.

$f'' (x) = \frac{- \frac{3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{| x + 6 |}}{{| x + 6 |}^{2}}$

Combine terms.

Rewrite $\frac{- \frac{3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{| x + 6 |}}{{| x + 6 |}^{2}}$ as a product.

$f'' (x) = - \frac{3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{| x + 6 |} \cdot \frac{1}{{| x + 6 |}^{2}}$

Multiply $\frac{1}{{| x + 6 |}^{2}}$ by $\frac{3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{| x + 6 |}$ .

$f'' (x) = - \frac{3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{{| x + 6 |}^{2} | x + 6 |}$

Multiply ${| x + 6 |}^{2}$ by $| x + 6 |$ by adding the exponents.

Multiply ${| x + 6 |}^{2}$ by $| x + 6 |$ .

Raise $| x + 6 |$ to the power of $1$ .

$f'' (x) = - \frac{3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{{| x + 6 |}^{2} | x + 6 |}$

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

$f'' (x) = - \frac{3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{{| x + 6 |}^{2 + 1}}$

Add $2$ and $1$ .

$f'' (x) = - \frac{3 (x^{2} + 12 x + 36 - | x^{2} + 12 x + 36 |)}{{| x + 6 |}^{3}}$

Step 3

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

$\frac{3 (x + 6)}{| x + 6 |} = 0$

Step 4

Find the first derivative.

By the Sum Rule, the derivative of $3 | x + 6 | + 2$ with respect to $x$ is $\frac{d}{d x} [3 | x + 6 |] + \frac{d}{d x} [2]$ .

$f' (x) = \frac{d}{d x} (3 | x + 6 |) + \frac{d}{d x} (2)$

Evaluate $\frac{d}{d x} [3 | x + 6 |]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 | x + 6 |$ with respect to $x$ is $3 \frac{d}{d x} [| x + 6 |]$ .

$f' (x) = 3 \frac{d}{d x} (| x + 6 |) + \frac{d}{d x} (2)$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = x + 6$ .

To apply the Chain Rule, set $u$ as $x + 6$ .

$f' (x) = 3 (\frac{d}{d u} (| u |) \frac{d}{d x} (x + 6)) + \frac{d}{d x} (2)$

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$f' (x) = 3 (\frac{u}{| u |} \cdot \frac{d}{d x} (x + 6)) + \frac{d}{d x} (2)$

Replace all occurrences of $u$ with $x + 6$ .

$f' (x) = 3 (\frac{x + 6}{| x + 6 |} \cdot \frac{d}{d x} (x + 6)) + \frac{d}{d x} (2)$

By the Sum Rule, the derivative of $x + 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [6]$ .

$f' (x) = 3 (\frac{x + 6}{| x + 6 |} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (6))) + \frac{d}{d x} (2)$

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

$f' (x) = 3 (\frac{x + 6}{| x + 6 |} \cdot (1 + \frac{d}{d x} (6))) + \frac{d}{d x} (2)$

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$f' (x) = 3 (\frac{x + 6}{| x + 6 |} \cdot (1 + 0)) + \frac{d}{d x} (2)$

Add $1$ and $0$ .

$f' (x) = 3 (\frac{x + 6}{| x + 6 |} \cdot 1) + \frac{d}{d x} (2)$

Multiply $\frac{x + 6}{| x + 6 |}$ by $1$ .

$f' (x) = 3 (\frac{x + 6}{| x + 6 |}) + \frac{d}{d x} (2)$

Combine $3$ and $\frac{x + 6}{| x + 6 |}$ .

$f' (x) = \frac{3 (x + 6)}{| x + 6 |} + \frac{d}{d x} (2)$

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

$f' (x) = \frac{3 (x + 6)}{| x + 6 |} + 0$

Simplify.

Apply the distributive property.

$f' (x) = \frac{3 x + 3 \cdot 6}{| x + 6 |} + 0$

Combine terms.

Multiply $3$ by $6$ .

$f' (x) = \frac{3 x + 18}{| x + 6 |} + 0$

Add $\frac{3 x + 18}{| x + 6 |}$ and $0$ .

$f' (x) = \frac{3 x + 18}{| x + 6 |}$

Factor $3$ out of $3 x + 18$ .

Factor $3$ out of $3 x$ .

$f' (x) = \frac{3 (x) + 18}{| x + 6 |}$

Factor $3$ out of $18$ .

$f' (x) = \frac{3 x + 3 \cdot 6}{| x + 6 |}$

Factor $3$ out of $3 x + 3 \cdot 6$ .

$f' (x) = \frac{3 (x + 6)}{| x + 6 |}$

The first derivative of $f (x)$ with respect to $x$ is $\frac{3 (x + 6)}{| x + 6 |}$ .

$\frac{3 (x + 6)}{| x + 6 |}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{3 (x + 6)}{| x + 6 |} = 0$ .

Set the first derivative equal to $0$ .

$\frac{3 (x + 6)}{| x + 6 |} = 0$

Set the numerator equal to zero.

$3 (x + 6) = 0$

Solve the equation for $x$ .

Divide each term in $3 (x + 6) = 0$ by $3$ and simplify.

Divide each term in $3 (x + 6) = 0$ by $3$ .

$\frac{3 (x + 6)}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (x + 6)}{3} = \frac{0}{3}$

Divide $x + 6$ by $1$ .

$x + 6 = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Exclude the solutions that do not make $\frac{3 (x + 6)}{| x + 6 |} = 0$ true.

$No solution$

Step 6

Find the values where the derivative is undefined.

Set the denominator in $\frac{3 (x + 6)}{| x + 6 |}$ equal to $0$ to find where the expression is undefined.

$| x + 6 | = 0$

Solve for $x$ .

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x + 6 = \pm 0$

Plus or minus $0$ is $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 7

Critical points to evaluate.

$x = - 6$

Step 8

Evaluate the second derivative at $x = - 6$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{3 ({(- 6)}^{2} + 12 (- 6) + 36 - | {(- 6)}^{2} + 12 (- 6) + 36 |)}{{| (- 6) + 6 |}^{3}}$

Step 9

Evaluate the second derivative.

Add $- 6$ and $6$ .

$- \frac{3 ({(- 6)}^{2} + 12 (- 6) + 36 - | {(- 6)}^{2} + 12 (- 6) + 36 |)}{{| 0 |}^{3}}$

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$- \frac{3 ({(- 6)}^{2} + 12 (- 6) + 36 - | {(- 6)}^{2} + 12 (- 6) + 36 |)}{0^{3}}$

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

$- \frac{3 ({(- 6)}^{2} + 12 (- 6) + 36 - | {(- 6)}^{2} + 12 (- 6) + 36 |)}{0}$

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

Undefined

Step 10

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - 6) \cup (- 6, \infty)$

Substitute any number, such as $- 8$ , from the interval $(- \infty, - 6)$ in the first derivative $\frac{3 (x + 6)}{| x + 6 |}$ to check if the result is negative or positive.

Replace the variable $x$ with $- 8$ in the expression.

$f' (- 8) = \frac{3 ((- 8) + 6)}{| (- 8) + 6 |}$

Simplify the result.

Add $- 8$ and $6$ .

$f' (- 8) = \frac{3 \cdot - 2}{| - 8 + 6 |}$

Simplify the denominator.

Add $- 8$ and $6$ .

$f' (- 8) = \frac{3 \cdot - 2}{| - 2 |}$

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f' (- 8) = \frac{3 \cdot - 2}{2}$

Simplify the expression.

Multiply $3$ by $- 2$ .

$f' (- 8) = \frac{- 6}{2}$

Divide $- 6$ by $2$ .

$f' (- 8) = - 3$

The final answer is $- 3$ .

$- 3$

Substitute any number, such as $0$ , from the interval $(- 6, \infty)$ in the first derivative $\frac{3 (x + 6)}{| x + 6 |}$ to check if the result is negative or positive.

Replace the variable $x$ with $0$ in the expression.

$f' (0) = \frac{3 ((0) + 6)}{| (0) + 6 |}$

Simplify the result.

Add $0$ and $6$ .

$f' (0) = \frac{3 \cdot 6}{| 0 + 6 |}$

Simplify the denominator.

Add $0$ and $6$ .

$f' (0) = \frac{3 \cdot 6}{| 6 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $6$ is $6$ .

$f' (0) = \frac{3 \cdot 6}{6}$

Simplify the expression.

Multiply $3$ by $6$ .

$f' (0) = \frac{18}{6}$

Divide $18$ by $6$ .

$f' (0) = 3$

The final answer is $3$ .

$3$

Since the first derivative changed signs from negative to positive around $x = - 6$ , then $x = - 6$ is a local minimum.

$x = - 6$ is a local minimum

Step 11

Find the Maximum/Minimum Value y=3|x+6|+2

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