Calculus Examples

$f (x) = x - x^{- 1}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x - x^{- 1}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- x^{- 1}]$ .

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

Step 1.1.1.2

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

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

Step 1.1.2

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

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

$1 - - x^{- 2}$

Step 1.1.2.3

Multiply $- 1$ by $- 1$ .

$1 + 1 x^{- 2}$

Step 1.1.2.4

Multiply $x^{- 2}$ by $1$ .

$1 + x^{- 2}$

Step 1.1.3

Reorder terms.

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

Step 1.2

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

$x^{- 2} + 1$

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

$x^{- 2} + 1 = 0$

Step 2.2

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

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

Step 2.3

Subtract $1$ from both sides of the equation.

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

Step 2.4

Find the LCD of the terms in the equation.

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Step 2.4.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x^{2}, 1$

Step 2.4.2

The LCM of one and any expression is the expression.

$x^{2}$

Step 2.5

Multiply each term in $\frac{1}{x^{2}} = - 1$ by $x^{2}$ to eliminate the fractions.

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Step 2.5.1

Multiply each term in $\frac{1}{x^{2}} = - 1$ by $x^{2}$ .

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

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Cancel the common factor of $x^{2}$ .

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Step 2.5.2.1.1

Cancel the common factor.

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

Step 2.5.2.1.2

Rewrite the expression.

$1 = - x^{2}$

Step 2.6

Solve the equation.

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Step 2.6.1

Rewrite the equation as $- x^{2} = 1$ .

$- x^{2} = 1$

Step 2.6.2

Divide each term in $- x^{2} = 1$ by $- 1$ and simplify.

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Step 2.6.2.1

Divide each term in $- x^{2} = 1$ by $- 1$ .

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

Step 2.6.2.2

Simplify the left side.

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Step 2.6.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.6.2.2.2

Divide $x^{2}$ by $1$ .

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

Step 2.6.2.3

Simplify the right side.

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Step 2.6.2.3.1

Divide $1$ by $- 1$ .

$x^{2} = - 1$

Step 2.6.3

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

$x = \pm \sqrt{- 1}$

Step 2.6.4

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 2.6.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.6.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Step 2.6.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

Step 2.6.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

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

Set the base in $x^{- 2}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 5

After finding the point that makes the derivative $f' (x) = x^{- 2} + 1$ equal to $0$ or undefined, the interval to check where $f (x) = x - x^{- 1}$ 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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Step 6.1

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

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

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

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

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

Step 6.2.1.2

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

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

Step 6.2.1.3

Divide $1$ by $1$ .

$f' (- 1) = 1 + 1$

Step 6.2.2

Add $1$ and $1$ .

$f' (- 1) = 2$

Step 6.2.3

The final answer is $2$ .

$2$

Step 6.3

At $x = - 1$ the derivative is $2$ . Since this is positive, the function is increasing on $(- \infty, 0)$ .

Increasing 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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Step 7.1

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

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

Step 7.2

Simplify the result.

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Step 7.2.1

One to any power is one.

$f' (1) = 1 + 1$

Step 7.2.2

Add $1$ and $1$ .

$f' (1) = 2$

Step 7.2.3

The final answer is $2$ .

$2$

Step 7.3

At $x = 1$ the derivative is $2$ . Since this is positive, the function is increasing on $(0, \infty)$ .

Increasing on $(0, \infty)$ since $f' (x) > 0$

Step 8

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

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

Step 9