Calculus Examples

$f (x) = \frac{1}{x}$ , $[2, 6]$

Step 1

Find the derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$\frac{d}{d x} [x^{- 1}]$

Step 1.1.2

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

$- x^{- 2}$

Step 1.1.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{1}{x^{2}}$ .

$- \frac{1}{x^{2}}$

Step 2

Find if the derivative is continuous on $[2, 6]$ .

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Step 2.1

To find whether the function is continuous on $[2, 6]$ or not, find the domain of $f' (x) = - \frac{1}{x^{2}}$ .

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Step 2.1.1

Set the denominator in $\frac{1}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 2.1.2

Solve for $x$ .

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Step 2.1.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 2.1.2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.1.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 2.1.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 2.1.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.1.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Step 2.2

$f' (x)$ is continuous on $[2, 6]$ .

The function is continuous.

Step 3

The function is differentiable on $[2, 6]$ because the derivative is continuous on $[2, 6]$ .

The function is differentiable.

Step 4

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