Calculus Examples

$f (x) = 4 x - 2$ , $(- 2, 4)$

Step 1

Check if $f (x) = 4 x - 2$ is continuous.

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Step 1.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 1.2

$f (x)$ is continuous on $[- 2, 4]$ .

The function is continuous.

Step 2

Check if $f (x) = 4 x - 2$ is differentiable.

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Step 2.1

Find the derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

By the Sum Rule, the derivative of $4 x - 2$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 2.1.1.2

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 2.1.1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 2.1.1.2.2

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

$4 \cdot 1 + \frac{d}{d x} [- 2]$

Step 2.1.1.2.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [- 2]$

Step 2.1.1.3

Differentiate using the Constant Rule.

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Step 2.1.1.3.1

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

$4 + 0$

Step 2.1.1.3.2

Add $4$ and $0$ .

$f' (x) = 4$

Step 2.1.2

The first derivative of $f (x)$ with respect to $x$ is $4$ .

$4$

Step 2.2

Find if the derivative is continuous on $[- 2, 4]$ .

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Step 2.2.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 2.2.2

$f' (x)$ is continuous on $[- 2, 4]$ .

The function is continuous.

Step 2.3

The function is differentiable on $[- 2, 4]$ because the derivative is continuous on $[- 2, 4]$ .

The function is differentiable.

Step 3

For arc length to be guaranteed, the function and its derivative must both be continuous on the closed interval $[- 2, 4]$ .

The function and its derivative are continuous on the closed interval $[- 2, 4]$ .

Step 4

Find the derivative of $f (x) = 4 x - 2$ .

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Step 4.1

By the Sum Rule, the derivative of $4 x - 2$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$

Step 4.2

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 4.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 4.2.2

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

$4 \cdot 1 + \frac{d}{d x} [- 2]$

Step 4.2.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [- 2]$

Step 4.3

Differentiate using the Constant Rule.

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Step 4.3.1

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

$4 + 0$

Step 4.3.2

Add $4$ and $0$ .

$4$

Step 5

To find the arc length of a function, use the formula $L = \int_{a}^{b} \sqrt{1 + {(f' (x))}^{2}} d x$ .

$\int_{- 2}^{4} \sqrt{1 + {(4)}^{2}} d x$

Step 6

Evaluate the integral.

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Step 6.1

Apply the constant rule.

$\sqrt{17} x]_{- 2}^{4}$

Step 6.2

Substitute and simplify.

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Step 6.2.1

Evaluate $\sqrt{17} x$ at $4$ and at $- 2$ .

$(\sqrt{17} \cdot 4) - \sqrt{17} \cdot - 2$

Step 6.2.2

Simplify.

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Step 6.2.2.1

Move $4$ to the left of $\sqrt{17}$ .

$4 \cdot \sqrt{17} - \sqrt{17} \cdot - 2$

Step 6.2.2.2

Multiply $- 2$ by $- 1$ .

$4 \sqrt{17} + 2 \sqrt{17}$

Step 6.2.2.3

Add $4 \sqrt{17}$ and $2 \sqrt{17}$ .

$6 \sqrt{17}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$6 \sqrt{17}$

Decimal Form:

$24.73863375 \dots$

Step 8

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