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

$y = | x + 3 | - 2$

Find the first derivative of the function.

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

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

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

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 $g (x) = x + 3$ .

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

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

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

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

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

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

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

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

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

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

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

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

Add $1$ and $0$ .

$\frac{x + 3}{| x + 3 |} \cdot 1 + \frac{d}{d x} [- 2]$

Multiply $\frac{x + 3}{| x + 3 |}$ by $1$ .

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

Differentiate using the Constant Rule.

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

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

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

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

Find the second derivative of the function.

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 $f (x) = x + 3$ and $g (x) = | x + 3 |$ .

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

Differentiate.

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

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

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

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

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

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

Simplify the expression.

Add $1$ and $0$ .

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

Multiply $| x + 3 |$ by $1$ .

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

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 $g (x) = x + 3$ .

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

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

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

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

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

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

Differentiate.

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

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

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

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

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

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

Simplify the expression.

Add $1$ and $0$ .

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

Multiply $\frac{x + 3}{| x + 3 |}$ by $1$ .

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

Simplify.

Apply the distributive property.

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

Simplify the numerator.

Simplify each term.

Multiply $- 1$ by $3$ .

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

Multiply $- x - 3$ by $\frac{x + 3}{| x + 3 |}$ .

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

Simplify the numerator.

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

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Factor $- 1$ out of $- (x) - 1 (3)$ .

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

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Move the negative in front of the fraction.

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $| x + 3 | | x + 3 |$ .

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Rewrite ${(x + 3)}^{2}$ as $(x + 3) (x + 3)$ .

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

Expand $(x + 3) (x + 3)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $3$ to the left of $x$ .

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

Multiply $3$ by $3$ .

$f'' (x) = \frac{\frac{| x^{2} + 3 x + 3 x + 9 | - {(x + 3)}^{2}}{| x + 3 |}}{{| x + 3 |}^{2}}$

Add $3 x$ and $3 x$ .

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - {(x + 3)}^{2}}{| x + 3 |}}{{| x + 3 |}^{2}}$

Rewrite ${(x + 3)}^{2}$ as $(x + 3) (x + 3)$ .

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - ((x + 3) (x + 3))}{| x + 3 |}}{{| x + 3 |}^{2}}$

Expand $(x + 3) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - (x (x + 3) + 3 (x + 3))}{| x + 3 |}}{{| x + 3 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - (x \cdot x + x \cdot 3 + 3 (x + 3))}{| x + 3 |}}{{| x + 3 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - (x \cdot x + x \cdot 3 + 3 x + 3 \cdot 3)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - (x^{2} + x \cdot 3 + 3 x + 3 \cdot 3)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Move $3$ to the left of $x$ .

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - (x^{2} + 3 \cdot x + 3 x + 3 \cdot 3)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Multiply $3$ by $3$ .

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - (x^{2} + 3 x + 3 x + 9)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Add $3 x$ and $3 x$ .

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - (x^{2} + 6 x + 9)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Apply the distributive property.

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - x^{2} - (6 x) - 1 \cdot 9}{| x + 3 |}}{{| x + 3 |}^{2}}$

Simplify.

Multiply $6$ by $- 1$ .

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - x^{2} - 6 x - 1 \cdot 9}{| x + 3 |}}{{| x + 3 |}^{2}}$

Multiply $- 1$ by $9$ .

$f'' (x) = \frac{\frac{| x^{2} + 6 x + 9 | - x^{2} - 6 x - 9}{| x + 3 |}}{{| x + 3 |}^{2}}$

Reorder terms.

$f'' (x) = \frac{\frac{- x^{2} - 6 x + | x^{2} + 6 x + 9 | - 9}{| x + 3 |}}{{| x + 3 |}^{2}}$

Rewrite $- x^{2} - 6 x + | x^{2} + 6 x + 9 | - 9$ in a factored form.

Regroup terms.

$f'' (x) = \frac{\frac{- 9 - x^{2} - 6 x + | x^{2} + 6 x + 9 |}{| x + 3 |}}{{| x + 3 |}^{2}}$

Factor $- 1$ out of $- x^{2} - 6 x + | x^{2} + 6 x + 9 |$ .

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

$f'' (x) = \frac{\frac{- 9 - (x^{2}) - 6 x + | x^{2} + 6 x + 9 |}{| x + 3 |}}{{| x + 3 |}^{2}}$

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

$f'' (x) = \frac{\frac{- 9 - (x^{2}) - (6 x) + | x^{2} + 6 x + 9 |}{| x + 3 |}}{{| x + 3 |}^{2}}$

Factor $- 1$ out of $| x^{2} + 6 x + 9 |$ .

$f'' (x) = \frac{\frac{- 9 - (x^{2}) - (6 x) - 1 (- | x^{2} + 6 x + 9 |)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Factor $- 1$ out of $- (x^{2}) - (6 x)$ .

$f'' (x) = \frac{\frac{- 9 - (x^{2} + 6 x) - 1 (- | x^{2} + 6 x + 9 |)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Factor $- 1$ out of $- (x^{2} + 6 x) - 1 (- | x^{2} + 6 x + 9 |)$ .

$f'' (x) = \frac{\frac{- 9 - (x^{2} + 6 x - | x^{2} + 6 x + 9 |)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Factor $- 1$ out of $- 9 - (x^{2} + 6 x - | x^{2} + 6 x + 9 |)$ .

Rewrite $- 9$ as $- 1 (9)$ .

$f'' (x) = \frac{\frac{- 1 \cdot 9 - (x^{2} + 6 x - | x^{2} + 6 x + 9 |)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Factor $- 1$ out of $- 1 (9) - (x^{2} + 6 x - | x^{2} + 6 x + 9 |)$ .

$f'' (x) = \frac{\frac{- 1 (9 + x^{2} + 6 x - | x^{2} + 6 x + 9 |)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Rewrite $- 1 (9 + x^{2} + 6 x - | x^{2} + 6 x + 9 |)$ as $- (9 + x^{2} + 6 x - | x^{2} + 6 x + 9 |)$ .

$f'' (x) = \frac{\frac{- (9 + x^{2} + 6 x - | x^{2} + 6 x + 9 |)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Reorder terms.

$f'' (x) = \frac{\frac{- (x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |)}{| x + 3 |}}{{| x + 3 |}^{2}}$

Move the negative in front of the fraction.

$f'' (x) = \frac{- \frac{x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |}{| x + 3 |}}{{| x + 3 |}^{2}}$

Combine terms.

Rewrite $\frac{- \frac{x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |}{| x + 3 |}}{{| x + 3 |}^{2}}$ as a product.

$f'' (x) = - \frac{x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |}{| x + 3 |} \cdot \frac{1}{{| x + 3 |}^{2}}$

Multiply $\frac{1}{{| x + 3 |}^{2}}$ by $\frac{x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |}{| x + 3 |}$ .

$f'' (x) = - \frac{x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |}{{| x + 3 |}^{2} | x + 3 |}$

Multiply ${| x + 3 |}^{2}$ by $| x + 3 |$ by adding the exponents.

Multiply ${| x + 3 |}^{2}$ by $| x + 3 |$ .

Raise $| x + 3 |$ to the power of $1$ .

$f'' (x) = - \frac{x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |}{{| x + 3 |}^{2} | x + 3 |}$

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

$f'' (x) = - \frac{x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |}{{| x + 3 |}^{2 + 1}}$

Add $2$ and $1$ .

$f'' (x) = - \frac{x^{2} + 6 x + 9 - | x^{2} + 6 x + 9 |}{{| x + 3 |}^{3}}$

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

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

Find the first derivative.

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

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

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

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 $g (x) = x + 3$ .

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

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

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

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

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

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

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

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

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

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

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

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

Add $1$ and $0$ .

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

Multiply $\frac{x + 3}{| x + 3 |}$ by $1$ .

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

Differentiate using the Constant Rule.

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

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

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

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

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

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

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

Set the first derivative equal to $0$ .

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

Set the numerator equal to zero.

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$No solution$

Find the values where the derivative is undefined.

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

$| x + 3 | = 0$

Solve for $x$ .

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

$x + 3 = \pm 0$

Plus or minus $0$ is $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Critical points to evaluate.

$x = - 3$

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

$- \frac{{(- 3)}^{2} + 6 (- 3) + 9 - | {(- 3)}^{2} + 6 (- 3) + 9 |}{{| (- 3) + 3 |}^{3}}$

Evaluate the second derivative.

Add $- 3$ and $3$ .

$- \frac{{(- 3)}^{2} + 6 (- 3) + 9 - | {(- 3)}^{2} + 6 (- 3) + 9 |}{{| 0 |}^{3}}$

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

$- \frac{{(- 3)}^{2} + 6 (- 3) + 9 - | {(- 3)}^{2} + 6 (- 3) + 9 |}{0^{3}}$

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

$- \frac{{(- 3)}^{2} + 6 (- 3) + 9 - | {(- 3)}^{2} + 6 (- 3) + 9 |}{0}$

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

Undefined

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.

$(- \infty, - 3) \cup (- 3, \infty)$

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

Replace the variable $x$ with $- 6$ in the expression.

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

Simplify the result.

Add $- 6$ and $3$ .

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

Simplify the denominator.

Add $- 6$ and $3$ .

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

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

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

Divide $- 3$ by $3$ .

$f' (- 6) = - 1$

The final answer is $- 1$ .

$- 1$

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

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

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

Simplify the result.

Add $0$ and $3$ .

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

Simplify the denominator.

Add $0$ and $3$ .

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

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

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

Divide $3$ by $3$ .

$f' (0) = 1$

The final answer is $1$ .

$1$

Since the first derivative changed signs from negative to positive around $x = - 3$ , then $x = - 3$ is a local minimum.

$x = - 3$ is a local minimum

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

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