Calculus Examples

$f (x) = \frac{1}{x}$

Step 1

Find the first derivative.

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Find the first derivative.

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

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

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

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

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

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

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

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

Step 2

Set the first derivative equal to $0$ then solve the equation $- \frac{1}{x^{2}} = 0$ .

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Set the first derivative equal to $0$ .

$- \frac{1}{x^{2}} = 0$

Set the numerator equal to zero.

$1 = 0$

Since $1 \neq 0$ , there are no solutions.

No solution

Step 3

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 4

Find where the derivative is undefined.

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

$x^{2} = 0$

Solve for $x$ .

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

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

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

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 5

After finding the point that makes the derivative $f' (x) = - \frac{1}{x^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{1}{x}$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

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

Step 6

Substitute a value from the interval $(- \infty, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $- 1$ in the expression.

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

Simplify the result.

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Raise $- 1$ to the power of $2$ .

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

Cancel the common factor of $1$ .

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Cancel the common factor.

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

Rewrite the expression.

$f' (- 1) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f' (- 1) = - 1$

The final answer is $- 1$ .

$- 1$

At $x = - 1$ the derivative is $- 1$ . Since this is negative, the function is decreasing on $(- \infty, 0)$ .

Decreasing on $(- \infty, 0)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Replace the variable $x$ with $1$ in the expression.

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

Simplify the result.

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One to any power is one.

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

Cancel the common factor of $1$ .

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Cancel the common factor.

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

Rewrite the expression.

$f' (1) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f' (1) = - 1$

The final answer is $- 1$ .

$- 1$

At $x = 1$ the derivative is $- 1$ . Since this is negative, the function is decreasing on $(0, \infty)$ .

Decreasing on $(0, \infty)$ since $f' (x) < 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Decreasing on: $(- \infty, 0), (0, \infty)$

Step 9