Calculus Examples

$f (x) = | x - 3 |$

Step 1

Find the first derivative of the function.

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Step 1.1

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 $g (x) = x - 3$ .

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

$\frac{x - 3}{| x - 3 |} (1 + 0)$

Step 1.2.4

Simplify the expression.

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Step 1.2.4.1

Add $1$ and $0$ .

$\frac{x - 3}{| x - 3 |} \cdot 1$

Step 1.2.4.2

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

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

Step 2

Find the second derivative of the function.

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Step 2.1

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 $g (x) = | x - 3 |$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Add $1$ and $0$ .

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

Step 2.2.4.2

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

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

Step 2.3

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 $g (x) = x - 3$ .

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

Differentiate.

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

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

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

Step 2.4.4

Simplify the expression.

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Step 2.4.4.1

Add $1$ and $0$ .

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

Step 2.4.4.2

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

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Simplify the numerator.

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Step 2.5.2.1

Simplify each term.

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Step 2.5.2.1.1

Multiply $- 1$ by $- 3$ .

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

Step 2.5.2.1.2

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

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

Step 2.5.2.1.3

Simplify the numerator.

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Step 2.5.2.1.3.1

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

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

Step 2.5.2.1.3.2

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

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

Step 2.5.2.1.3.3

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

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

Step 2.5.2.1.3.4

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

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

Step 2.5.2.1.3.5

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

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

Step 2.5.2.1.3.6

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

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

Step 2.5.2.1.3.7

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

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

Step 2.5.2.1.3.8

Add $1$ and $1$ .

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

Step 2.5.2.1.4

Move the negative in front of the fraction.

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

Step 2.5.2.2

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

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

Step 2.5.2.3

Combine the numerators over the common denominator.

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

Step 2.5.2.4

Simplify the numerator.

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Step 2.5.2.4.1

Multiply $| x - 3 | | x - 3 |$ .

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Step 2.5.2.4.1.1

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

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

Step 2.5.2.4.1.2

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

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

Step 2.5.2.4.1.3

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

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

Step 2.5.2.4.1.4

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

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

Step 2.5.2.4.1.5

Add $1$ and $1$ .

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

Step 2.5.2.4.2

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

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

Step 2.5.2.4.3

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

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Step 2.5.2.4.3.1

Apply the distributive property.

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

Step 2.5.2.4.3.2

Apply the distributive property.

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

Step 2.5.2.4.3.3

Apply the distributive property.

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

Step 2.5.2.4.4

Simplify and combine like terms.

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Step 2.5.2.4.4.1

Simplify each term.

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Step 2.5.2.4.4.1.1

Multiply $x$ by $x$ .

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

Step 2.5.2.4.4.1.2

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

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

Step 2.5.2.4.4.1.3

Multiply $- 3$ by $- 3$ .

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

Step 2.5.2.4.4.2

Subtract $3 x$ from $- 3 x$ .

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

Step 2.5.2.4.5

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

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

Step 2.5.2.4.6

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

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Step 2.5.2.4.6.1

Apply the distributive property.

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

Step 2.5.2.4.6.2

Apply the distributive property.

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

Step 2.5.2.4.6.3

Apply the distributive property.

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

Step 2.5.2.4.7

Simplify and combine like terms.

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Step 2.5.2.4.7.1

Simplify each term.

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Step 2.5.2.4.7.1.1

Multiply $x$ by $x$ .

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

Step 2.5.2.4.7.1.2

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

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

Step 2.5.2.4.7.1.3

Multiply $- 3$ by $- 3$ .

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

Step 2.5.2.4.7.2

Subtract $3 x$ from $- 3 x$ .

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

Step 2.5.2.4.8

Apply the distributive property.

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

Step 2.5.2.4.9

Simplify.

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Step 2.5.2.4.9.1

Multiply $- 6$ by $- 1$ .

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

Step 2.5.2.4.9.2

Multiply $- 1$ by $9$ .

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

Step 2.5.2.4.10

Reorder terms.

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

Step 2.5.2.4.11

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

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Step 2.5.2.4.11.1

Regroup terms.

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

Step 2.5.2.4.11.2

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

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Step 2.5.2.4.11.2.1

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

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

Step 2.5.2.4.11.2.2

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

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

Step 2.5.2.4.11.2.3

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

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

Step 2.5.2.4.11.2.4

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

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

Step 2.5.2.4.11.2.5

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

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

Step 2.5.2.4.11.3

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

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Step 2.5.2.4.11.3.1

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

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

Step 2.5.2.4.11.3.2

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

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

Step 2.5.2.4.11.3.3

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

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

Step 2.5.2.4.11.4

Reorder terms.

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

Step 2.5.2.5

Move the negative in front of the fraction.

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

Step 2.5.3

Combine terms.

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Step 2.5.3.1

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

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

Step 2.5.3.2

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

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

Step 2.5.3.3

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

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Step 2.5.3.3.1

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

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Step 2.5.3.3.1.1

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

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

Step 2.5.3.3.1.2

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

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

Step 2.5.3.3.2

Add $2$ and $1$ .

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

Step 3

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

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

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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 $g (x) = x - 3$ .

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Step 4.1.1.1

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

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

Step 4.1.1.2

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

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

Step 4.1.1.3

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

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

Step 4.1.2

Differentiate.

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Step 4.1.2.1

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

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

Step 4.1.2.2

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

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

Step 4.1.2.3

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

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

Step 4.1.2.4

Simplify the expression.

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Step 4.1.2.4.1

Add $1$ and $0$ .

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

Step 4.1.2.4.2

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

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

Step 4.2

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

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

Step 5

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

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

Set the numerator equal to zero.

$x - 3 = 0$

Step 5.3

Add $3$ to both sides of the equation.

$x = 3$

Step 5.4

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

$No solution$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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

$| x - 3 | = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

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

$x - 3 = \pm 0$

Step 6.2.2

Plus or minus $0$ is $0$ .

$x - 3 = 0$

Step 6.2.3

Add $3$ to both sides of the equation.

$x = 3$

Step 7

Critical points to evaluate.

$x = 3$

Step 8

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.

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

Step 9

Evaluate the second derivative.

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Step 9.1

Subtract $3$ from $3$ .

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

Step 9.2

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

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

Step 9.3

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

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

Step 9.4

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.

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Step 10.1

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

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

Step 10.2

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

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Step 10.2.1

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

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

Step 10.2.2

Simplify the result.

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Step 10.2.2.1

Subtract $3$ from $0$ .

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

Step 10.2.2.2

Simplify the denominator.

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Step 10.2.2.2.1

Subtract $3$ from $0$ .

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

Step 10.2.2.2.2

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

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

Step 10.2.2.3

Divide $- 3$ by $3$ .

$f' (0) = - 1$

Step 10.2.2.4

The final answer is $- 1$ .

$- 1$

Step 10.3

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

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Step 10.3.1

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

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

Step 10.3.2

Simplify the result.

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Step 10.3.2.1

Subtract $3$ from $6$ .

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

Step 10.3.2.2

Simplify the denominator.

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Step 10.3.2.2.1

Subtract $3$ from $6$ .

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

Step 10.3.2.2.2

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

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

Step 10.3.2.3

Divide $3$ by $3$ .

$f' (6) = 1$

Step 10.3.2.4

The final answer is $1$ .

$1$

Step 10.4

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

Step 11