Calculus Examples

$y = 2 - | t - 2 |$ , $[- 9, 4]$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate.

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Step 1.1.1.1.1

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

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

Step 1.1.1.1.2

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

$0 + \frac{d}{d t} [- | t - 2 |]$

Step 1.1.1.2

Evaluate $\frac{d}{d t} [- | t - 2 |]$ .

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Step 1.1.1.2.1

Since $- 1$ is constant with respect to $t$ , the derivative of $- | t - 2 |$ with respect to $t$ is $- \frac{d}{d t} [| t - 2 |]$ .

$0 - \frac{d}{d t} [| t - 2 |]$

Step 1.1.1.2.2

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

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Step 1.1.1.2.2.1

To apply the Chain Rule, set $u$ as $t - 2$ .

$0 - (\frac{d}{d u} [| u |] \frac{d}{d t} [t - 2])$

Step 1.1.1.2.2.2

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

$0 - (\frac{u}{| u |} \frac{d}{d t} [t - 2])$

Step 1.1.1.2.2.3

Replace all occurrences of $u$ with $t - 2$ .

$0 - (\frac{t - 2}{| t - 2 |} \frac{d}{d t} [t - 2])$

Step 1.1.1.2.3

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

$0 - (\frac{t - 2}{| t - 2 |} (\frac{d}{d t} [t] + \frac{d}{d t} [- 2]))$

Step 1.1.1.2.4

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

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

Step 1.1.1.2.5

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

$0 - (\frac{t - 2}{| t - 2 |} (1 + 0))$

Step 1.1.1.2.6

Add $1$ and $0$ .

$0 - (\frac{t - 2}{| t - 2 |} \cdot 1)$

Step 1.1.1.2.7

Multiply $\frac{t - 2}{| t - 2 |}$ by $1$ .

$0 - \frac{t - 2}{| t - 2 |}$

Step 1.1.1.3

Subtract $\frac{t - 2}{| t - 2 |}$ from $0$ .

$f' (t) = - \frac{t - 2}{| t - 2 |}$

Step 1.1.2

The first derivative of $f (t)$ with respect to $t$ is $- \frac{t - 2}{| t - 2 |}$ .

$- \frac{t - 2}{| t - 2 |}$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $- \frac{t - 2}{| t - 2 |} = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$- \frac{t - 2}{| t - 2 |} = 0$

Step 1.2.2

Set the numerator equal to zero.

$t - 2 = 0$

Step 1.2.3

Add $2$ to both sides of the equation.

$t = 2$

Step 1.2.4

Exclude the solutions that do not make $- \frac{t - 2}{| t - 2 |} = 0$ true.

$No solution$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

Set the denominator in $\frac{t - 2}{| t - 2 |}$ equal to $0$ to find where the expression is undefined.

$| t - 2 | = 0$

Step 1.3.2

Solve for $t$ .

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Step 1.3.2.1

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

$t - 2 = \pm 0$

Step 1.3.2.2

Plus or minus $0$ is $0$ .

$t - 2 = 0$

Step 1.3.2.3

Add $2$ to both sides of the equation.

$t = 2$

Step 1.4

Evaluate $2 - | t - 2 |$ at each $t$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $t = 2$ .

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Step 1.4.1.1

Substitute $2$ for $t$ .

$2 - | (2) - 2 |$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify each term.

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Step 1.4.1.2.1.1

Subtract $2$ from $2$ .

$2 - | 0 |$

Step 1.4.1.2.1.2

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

$2 - 0$

Step 1.4.1.2.1.3

Multiply $- 1$ by $0$ .

$2 + 0$

Step 1.4.1.2.2

Add $2$ and $0$ .

$2$

Step 1.4.2

List all of the points.

$(2, 2)$

Step 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at $t = - 9$ .

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Step 2.1.1

Substitute $- 9$ for $t$ .

$2 - | (- 9) - 2 |$

Step 2.1.2

Simplify.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

Subtract $2$ from $- 9$ .

$2 - | - 11 |$

Step 2.1.2.1.2

The absolute value is the distance between a number and zero. The distance between $- 11$ and $0$ is $11$ .

$2 - 1 \cdot 11$

Step 2.1.2.1.3

Multiply $- 1$ by $11$ .

$2 - 11$

Step 2.1.2.2

Subtract $11$ from $2$ .

$- 9$

Step 2.2

Evaluate at $t = 4$ .

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Step 2.2.1

Substitute $4$ for $t$ .

$2 - | (4) - 2 |$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

Subtract $2$ from $4$ .

$2 - | 2 |$

Step 2.2.2.1.2

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

$2 - 1 \cdot 2$

Step 2.2.2.1.3

Multiply $- 1$ by $2$ .

$2 - 2$

Step 2.2.2.2

Subtract $2$ from $2$ .

$0$

Step 2.3

List all of the points.

$(- 9, - 9), (4, 0)$

Step 3

Compare the $f (t)$ values found for each value of $t$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (t)$ value and the minimum will occur at the lowest $f (t)$ value.

Absolute Maximum: $(2, 2)$

Absolute Minimum: $(- 9, - 9)$

Step 4