Calculus Examples

f (x) = x + \frac{4}{x}

$f (x) = x + \frac{4}{x}$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of

$x + \frac{4}{x}$ with respect to

$x$ is

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

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

Step 1.1.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 1.2

Evaluate

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

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Step 1.2.1

Since

$4$ is constant with respect to

$x$ , the derivative of

$\frac{4}{x}$ with respect to

$x$ is

$4 \frac{d}{d x} [\frac{1}{x}]$ .

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

Step 1.2.2

Rewrite

$\frac{1}{x}$ as

$x^{- 1}$ .

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

Step 1.2.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = - 1$ .

$1 + 4 (- x^{- 2})$

Step 1.2.4

Multiply

$- 1$ by

$4$ .

$1 - 4 x^{- 2}$

Step 1.3

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$1 - 4 \frac{1}{x^{2}}$

Step 1.4

Simplify.

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Step 1.4.1

Combine terms.

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Step 1.4.1.1

Combine

$- 4$ and

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

$1 + \frac{- 4}{x^{2}}$

Step 1.4.1.2

Move the negative in front of the fraction.

$1 - \frac{4}{x^{2}}$

Step 1.4.2

Reorder terms.

$- \frac{4}{x^{2}} + 1$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of

$- \frac{4}{x^{2}} + 1$ with respect to

$x$ is

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

$f'' (x) = \frac{d}{d x} (- \frac{4}{x^{2}}) + \frac{d}{d x} (1)$

Step 2.2

Evaluate

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

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Step 2.2.1

Since

$- 4$ is constant with respect to

$x$ , the derivative of

$- \frac{4}{x^{2}}$ with respect to

$x$ is

$- 4 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$f'' (x) = - 4 \frac{d}{d x} \frac{1}{x^{2}} + \frac{d}{d x} (1)$

Step 2.2.2

Rewrite

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

${(x^{2})}^{- 1}$ .

$f'' (x) = - 4 \frac{d}{d x} {(x^{2})}^{- 1} + \frac{d}{d x} (1)$

Step 2.2.3

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

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

$g (x) = x^{2}$ .

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Step 2.2.3.1

To apply the Chain Rule, set

$u$ as

$x^{2}$ .

$f'' (x) = - 4 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{2})) + \frac{d}{d x} (1)$

Step 2.2.3.2

Differentiate using the Power Rule which states that

$\frac{d}{d u} [u^{n}]$ is

$n u^{n - 1}$ where

$n = - 1$ .

$f'' (x) = - 4 (- u^{- 2} \frac{d}{d x} x^{2}) + \frac{d}{d x} (1)$

Step 2.2.3.3

Replace all occurrences of

$u$ with

$x^{2}$ .

$f'' (x) = - 4 (- {(x^{2})}^{- 2} \frac{d}{d x} x^{2}) + \frac{d}{d x} (1)$

Step 2.2.4

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 2$ .

$f'' (x) = - 4 (- {(x^{2})}^{- 2} (2 x)) + \frac{d}{d x} (1)$

Step 2.2.5

Multiply the exponents in

${(x^{2})}^{- 2}$ .

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Step 2.2.5.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = - 4 (- x^{2 \cdot - 2} (2 x)) + \frac{d}{d x} (1)$

Step 2.2.5.2

Multiply

$2$ by

$- 2$ .

$f'' (x) = - 4 (- x^{- 4} (2 x)) + \frac{d}{d x} (1)$

Step 2.2.6

Multiply

$2$ by

$- 1$ .

$f'' (x) = - 4 (- 2 x^{- 4} x) + \frac{d}{d x} (1)$

Step 2.2.7

Raise

$x$ to the power of

$1$ .

$f'' (x) = - 4 (- 2 (x \cdot x^{- 4})) + \frac{d}{d x} (1)$

Step 2.2.8

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - 4 (- 2 x^{1 - 4}) + \frac{d}{d x} (1)$

Step 2.2.9

Subtract

$4$ from

$1$ .

$f'' (x) = - 4 (- 2 x^{- 3}) + \frac{d}{d x} (1)$

Step 2.2.10

Multiply

$- 2$ by

$- 4$ .

$f'' (x) = 8 x^{- 3} + \frac{d}{d x} (1)$

Step 2.3

Since

$1$ is constant with respect to

$x$ , the derivative of

$1$ with respect to

$x$ is

$0$ .

$f'' (x) = 8 x^{- 3} + 0$

Step 2.4

Simplify.

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Step 2.4.1

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = 8 (\frac{1}{x^{3}}) + 0$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine

$8$ and

$\frac{1}{x^{3}}$ .

$f'' (x) = \frac{8}{x^{3}} + 0$

Step 2.4.2.2

Add

$\frac{8}{x^{3}}$ and

$0$ .

$f'' (x) = \frac{8}{x^{3}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to

$0$ and solve.

$- \frac{4}{x^{2}} + 1 = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Differentiate.

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Step 4.1.1.1

By the Sum Rule, the derivative of

$x + \frac{4}{x}$ with respect to

$x$ is

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

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

Step 4.1.1.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 4.1.2

Evaluate

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

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Step 4.1.2.1

Since

$4$ is constant with respect to

$x$ , the derivative of

$\frac{4}{x}$ with respect to

$x$ is

$4 \frac{d}{d x} [\frac{1}{x}]$ .

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

Step 4.1.2.2

Rewrite

$\frac{1}{x}$ as

$x^{- 1}$ .

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

Step 4.1.2.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = - 1$ .

$1 + 4 (- x^{- 2})$

Step 4.1.2.4

Multiply

$- 1$ by

$4$ .

$1 - 4 x^{- 2}$

Step 4.1.3

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$1 - 4 \frac{1}{x^{2}}$

Step 4.1.4

Simplify.

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Step 4.1.4.1

Combine terms.

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Step 4.1.4.1.1

Combine

$- 4$ and

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

$1 + \frac{- 4}{x^{2}}$

Step 4.1.4.1.2

Move the negative in front of the fraction.

$1 - \frac{4}{x^{2}}$

Step 4.1.4.2

Reorder terms.

$f' (x) = - \frac{4}{x^{2}} + 1$

Step 4.2

The first derivative of

$f (x)$ with respect to

$x$ is

$- \frac{4}{x^{2}} + 1$ .

$- \frac{4}{x^{2}} + 1$

Step 5

Set the first derivative equal to

$0$ then solve the equation

$- \frac{4}{x^{2}} + 1 = 0$ .

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Step 5.1

Set the first derivative equal to

$0$ .

$- \frac{4}{x^{2}} + 1 = 0$

Step 5.2

Subtract

$1$ from both sides of the equation.

$- \frac{4}{x^{2}} = - 1$

Step 5.3

Find the LCD of the terms in the equation.

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Step 5.3.1

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 5.3.2

The LCM of one and any expression is the expression.

$x^{2}$

Step 5.4

Multiply each term in

$- \frac{4}{x^{2}} = - 1$ by

$x^{2}$ to eliminate the fractions.

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Step 5.4.1

Multiply each term in

$- \frac{4}{x^{2}} = - 1$ by

$x^{2}$ .

$- \frac{4}{x^{2}} x^{2} = - x^{2}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of

$x^{2}$ .

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Step 5.4.2.1.1

Move the leading negative in

$- \frac{4}{x^{2}}$ into the numerator.

$\frac{- 4}{x^{2}} x^{2} = - x^{2}$

Step 5.4.2.1.2

Cancel the common factor.

$\frac{- 4}{x^{2}} x^{2} = - x^{2}$

Step 5.4.2.1.3

Rewrite the expression.

$- 4 = - x^{2}$

Step 5.5

Solve the equation.

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Step 5.5.1

Rewrite the equation as

$- x^{2} = - 4$ .

$- x^{2} = - 4$

Step 5.5.2

Divide each term in

$- x^{2} = - 4$ by

$- 1$ and simplify.

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Step 5.5.2.1

Divide each term in

$- x^{2} = - 4$ by

$- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 4}{- 1}$

Step 5.5.2.2

Simplify the left side.

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Step 5.5.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 4}{- 1}$

Step 5.5.2.2.2

Divide

$x^{2}$ by

$1$ .

$x^{2} = \frac{- 4}{- 1}$

Step 5.5.2.3

Simplify the right side.

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Step 5.5.2.3.1

Divide

$- 4$ by

$- 1$ .

$x^{2} = 4$

Step 5.5.3

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

$x = \pm \sqrt{4}$

Step 5.5.4

Simplify

$\pm \sqrt{4}$ .

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Step 5.5.4.1

Rewrite

$4$ as

$2^{2}$ .

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

Step 5.5.4.2

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

$x = \pm 2$

Step 5.5.5

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

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Step 5.5.5.1

First, use the positive value of the

$\pm$ to find the first solution.

$x = 2$

Step 5.5.5.2

Next, use the negative value of the

$\pm$ to find the second solution.

$x = - 2$

Step 5.5.5.3

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

$x = 2, - 2$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in

$\frac{4}{x^{2}}$ equal to

$0$ to find where the expression is undefined.

$x^{2} = 0$

Step 6.2

Solve for

$x$ .

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Step 6.2.1

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

$x = \pm \sqrt{0}$

Step 6.2.2

Simplify

$\pm \sqrt{0}$ .

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Step 6.2.2.1

Rewrite

$0$ as

$0^{2}$ .

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

Step 6.2.2.2

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

$x = \pm 0$

Step 6.2.2.3

Plus or minus

$0$ is

$0$ .

$x = 0$

Step 7

Critical points to evaluate.

$x = 2, - 2$

Step 8

Evaluate the second derivative at

$x = 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{8}{{(2)}^{3}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Raise

$2$ to the power of

$3$ .

$\frac{8}{8}$

Step 9.2

Divide

$8$ by

$8$ .

$1$

Step 10

$x = 2$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 2$ is a local minimum

Step 11

Find the y-value when

$x = 2$ .

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Step 11.1

Replace the variable

$x$ with

$2$ in the expression.

$f (2) = (2) + \frac{4}{2}$

Step 11.2

Simplify the result.

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Step 11.2.1

Divide

$4$ by

$2$ .

$f (2) = 2 + 2$

Step 11.2.2

Add

$2$ and

$2$ .

$f (2) = 4$

Step 11.2.3

The final answer is

$4$ .

$y = 4$

Step 12

Evaluate the second derivative at

$x = - 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{8}{{(- 2)}^{3}}$

Step 13

Evaluate the second derivative.

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Step 13.1

Raise

$- 2$ to the power of

$3$ .

$\frac{8}{- 8}$

Step 13.2

Divide

$8$ by

$- 8$ .

$- 1$

Step 14

$x = - 2$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 2$ is a local maximum

Step 15

Find the y-value when

$x = - 2$ .

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Step 15.1

Replace the variable

$x$ with

$- 2$ in the expression.

$f (- 2) = (- 2) + \frac{4}{- 2}$

Step 15.2

Simplify the result.

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Step 15.2.1

Divide

$4$ by

$- 2$ .

$f (- 2) = - 2 - 2$

Step 15.2.2

Subtract

$2$ from

$- 2$ .

$f (- 2) = - 4$

Step 15.2.3

The final answer is

$- 4$ .

$y = - 4$

Step 16

These are the local extrema for

$f (x) = x + \frac{4}{x}$ .

$(2, 4)$ is a local minima

$(- 2, - 4)$ is a local maxima

Step 17