Calculus Examples

$x + \frac{25}{x}$

Step 1

Write $x + \frac{25}{x}$ as a function.

$f (x) = x + \frac{25}{x}$

Step 2

Find the first derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $x + \frac{25}{x}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\frac{25}{x}]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [\frac{25}{x}]$

Step 2.1.2

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

$1 + \frac{d}{d x} [\frac{25}{x}]$

Step 2.2

Evaluate $\frac{d}{d x} [\frac{25}{x}]$ .

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Step 2.2.1

Since $25$ is constant with respect to $x$ , the derivative of $\frac{25}{x}$ with respect to $x$ is $25 \frac{d}{d x} [\frac{1}{x}]$ .

$1 + 25 \frac{d}{d x} [\frac{1}{x}]$

Step 2.2.2

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

$1 + 25 \frac{d}{d x} [x^{- 1}]$

Step 2.2.3

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

$1 + 25 (- x^{- 2})$

Step 2.2.4

Multiply $- 1$ by $25$ .

$1 - 25 x^{- 2}$

Step 2.3

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

$1 - 25 \frac{1}{x^{2}}$

Step 2.4

Simplify.

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Step 2.4.1

Combine terms.

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Step 2.4.1.1

Combine $- 25$ and $\frac{1}{x^{2}}$ .

$1 + \frac{- 25}{x^{2}}$

Step 2.4.1.2

Move the negative in front of the fraction.

$1 - \frac{25}{x^{2}}$

Step 2.4.2

Reorder terms.

$- \frac{25}{x^{2}} + 1$

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $- \frac{25}{x^{2}} + 1$ with respect to $x$ is $\frac{d}{d x} [- \frac{25}{x^{2}}] + \frac{d}{d x} [1]$ .

$f'' (x) = \frac{d}{d x} (- \frac{25}{x^{2}}) + \frac{d}{d x} (1)$

Step 3.2

Evaluate $\frac{d}{d x} [- \frac{25}{x^{2}}]$ .

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Step 3.2.1

Since $- 25$ is constant with respect to $x$ , the derivative of $- \frac{25}{x^{2}}$ with respect to $x$ is $- 25 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$f'' (x) = - 25 \frac{d}{d x} \frac{1}{x^{2}} + \frac{d}{d x} (1)$

Step 3.2.2

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$f'' (x) = - 25 \frac{d}{d x} {(x^{2})}^{- 1} + \frac{d}{d x} (1)$

Step 3.2.3

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 3.2.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

$f'' (x) = - 25 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{2})) + \frac{d}{d x} (1)$

Step 3.2.3.2

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

$f'' (x) = - 25 (- u^{- 2} \frac{d}{d x} x^{2}) + \frac{d}{d x} (1)$

Step 3.2.3.3

Replace all occurrences of $u$ with $x^{2}$ .

$f'' (x) = - 25 (- {(x^{2})}^{- 2} \frac{d}{d x} x^{2}) + \frac{d}{d x} (1)$

Step 3.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = - 25 (- {(x^{2})}^{- 2} (2 x)) + \frac{d}{d x} (1)$

Step 3.2.5

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 3.2.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = - 25 (- x^{2 \cdot - 2} (2 x)) + \frac{d}{d x} (1)$

Step 3.2.5.2

Multiply $2$ by $- 2$ .

$f'' (x) = - 25 (- x^{- 4} (2 x)) + \frac{d}{d x} (1)$

Step 3.2.6

Multiply $2$ by $- 1$ .

$f'' (x) = - 25 (- 2 x^{- 4} x) + \frac{d}{d x} (1)$

Step 3.2.7

Raise $x$ to the power of $1$ .

$f'' (x) = - 25 (- 2 (x \cdot x^{- 4})) + \frac{d}{d x} (1)$

Step 3.2.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - 25 (- 2 x^{1 - 4}) + \frac{d}{d x} (1)$

Step 3.2.9

Subtract $4$ from $1$ .

$f'' (x) = - 25 (- 2 x^{- 3}) + \frac{d}{d x} (1)$

Step 3.2.10

Multiply $- 2$ by $- 25$ .

$f'' (x) = 50 x^{- 3} + \frac{d}{d x} (1)$

Step 3.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = 50 x^{- 3} + 0$

Step 3.4

Simplify.

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Step 3.4.1

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

$f'' (x) = 50 (\frac{1}{x^{3}}) + 0$

Step 3.4.2

Combine terms.

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Step 3.4.2.1

Combine $50$ and $\frac{1}{x^{3}}$ .

$f'' (x) = \frac{50}{x^{3}} + 0$

Step 3.4.2.2

Add $\frac{50}{x^{3}}$ and $0$ .

$f'' (x) = \frac{50}{x^{3}}$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- \frac{25}{x^{2}} + 1 = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

Differentiate.

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Step 5.1.1.1

By the Sum Rule, the derivative of $x + \frac{25}{x}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\frac{25}{x}]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [\frac{25}{x}]$

Step 5.1.1.2

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

$1 + \frac{d}{d x} [\frac{25}{x}]$

Step 5.1.2

Evaluate $\frac{d}{d x} [\frac{25}{x}]$ .

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Step 5.1.2.1

Since $25$ is constant with respect to $x$ , the derivative of $\frac{25}{x}$ with respect to $x$ is $25 \frac{d}{d x} [\frac{1}{x}]$ .

$1 + 25 \frac{d}{d x} [\frac{1}{x}]$

Step 5.1.2.2

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

$1 + 25 \frac{d}{d x} [x^{- 1}]$

Step 5.1.2.3

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

$1 + 25 (- x^{- 2})$

Step 5.1.2.4

Multiply $- 1$ by $25$ .

$1 - 25 x^{- 2}$

Step 5.1.3

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

$1 - 25 \frac{1}{x^{2}}$

Step 5.1.4

Simplify.

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Step 5.1.4.1

Combine terms.

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Step 5.1.4.1.1

Combine $- 25$ and $\frac{1}{x^{2}}$ .

$1 + \frac{- 25}{x^{2}}$

Step 5.1.4.1.2

Move the negative in front of the fraction.

$1 - \frac{25}{x^{2}}$

Step 5.1.4.2

Reorder terms.

$f' (x) = - \frac{25}{x^{2}} + 1$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{25}{x^{2}} + 1$ .

$- \frac{25}{x^{2}} + 1$

Step 6

Set the first derivative equal to $0$ then solve the equation $- \frac{25}{x^{2}} + 1 = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$- \frac{25}{x^{2}} + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$- \frac{25}{x^{2}} = - 1$

Step 6.3

Find the LCD of the terms in the equation.

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Step 6.3.1

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 6.3.2

The LCM of one and any expression is the expression.

$x^{2}$

Step 6.4

Multiply each term in $- \frac{25}{x^{2}} = - 1$ by $x^{2}$ to eliminate the fractions.

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Step 6.4.1

Multiply each term in $- \frac{25}{x^{2}} = - 1$ by $x^{2}$ .

$- \frac{25}{x^{2}} x^{2} = - x^{2}$

Step 6.4.2

Simplify the left side.

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Step 6.4.2.1

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

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Step 6.4.2.1.1

Move the leading negative in $- \frac{25}{x^{2}}$ into the numerator.

$\frac{- 25}{x^{2}} x^{2} = - x^{2}$

Step 6.4.2.1.2

Cancel the common factor.

$\frac{- 25}{x^{2}} x^{2} = - x^{2}$

Step 6.4.2.1.3

Rewrite the expression.

$- 25 = - x^{2}$

Step 6.5

Solve the equation.

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Step 6.5.1

Rewrite the equation as $- x^{2} = - 25$ .

$- x^{2} = - 25$

Step 6.5.2

Divide each term in $- x^{2} = - 25$ by $- 1$ and simplify.

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Step 6.5.2.1

Divide each term in $- x^{2} = - 25$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 25}{- 1}$

Step 6.5.2.2

Simplify the left side.

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Step 6.5.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 25}{- 1}$

Step 6.5.2.2.2

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

$x^{2} = \frac{- 25}{- 1}$

Step 6.5.2.3

Simplify the right side.

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Step 6.5.2.3.1

Divide $- 25$ by $- 1$ .

$x^{2} = 25$

Step 6.5.3

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

$x = \pm \sqrt{25}$

Step 6.5.4

Simplify $\pm \sqrt{25}$ .

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Step 6.5.4.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 6.5.4.2

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

$x = \pm 5$

Step 6.5.5

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

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Step 6.5.5.1

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

$x = 5$

Step 6.5.5.2

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

$x = - 5$

Step 6.5.5.3

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

$x = 5, - 5$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

Set the denominator in $\frac{25}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 7.2

Solve for $x$ .

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Step 7.2.1

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

$x = \pm \sqrt{0}$

Step 7.2.2

Simplify $\pm \sqrt{0}$ .

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Step 7.2.2.1

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

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

Step 7.2.2.2

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

$x = \pm 0$

Step 7.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 8

Critical points to evaluate.

$x = 5, - 5$

Step 9

Evaluate the second derivative at $x = 5$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{50}{{(5)}^{3}}$

Step 10

Evaluate the second derivative.

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Step 10.1

Raise $5$ to the power of $3$ .

$\frac{50}{125}$

Step 10.2

Cancel the common factor of $50$ and $125$ .

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Step 10.2.1

Factor $25$ out of $50$ .

$\frac{25 (2)}{125}$

Step 10.2.2

Cancel the common factors.

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Step 10.2.2.1

Factor $25$ out of $125$ .

$\frac{25 \cdot 2}{25 \cdot 5}$

Step 10.2.2.2

Cancel the common factor.

$\frac{25 \cdot 2}{25 \cdot 5}$

Step 10.2.2.3

Rewrite the expression.

$\frac{2}{5}$

Step 11

$x = 5$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 5$ is a local minimum

Step 12

Find the y-value when $x = 5$ .

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Step 12.1

Replace the variable $x$ with $5$ in the expression.

$f (5) = (5) + \frac{25}{5}$

Step 12.2

Simplify the result.

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Step 12.2.1

Divide $25$ by $5$ .

$f (5) = 5 + 5$

Step 12.2.2

Add $5$ and $5$ .

$f (5) = 10$

Step 12.2.3

The final answer is $10$ .

$y = 10$

Step 13

Evaluate the second derivative at $x = - 5$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{50}{{(- 5)}^{3}}$

Step 14

Evaluate the second derivative.

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Step 14.1

Raise $- 5$ to the power of $3$ .

$\frac{50}{- 125}$

Step 14.2

Cancel the common factor of $50$ and $- 125$ .

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Step 14.2.1

Factor $25$ out of $50$ .

$\frac{25 (2)}{- 125}$

Step 14.2.2

Cancel the common factors.

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Step 14.2.2.1

Factor $25$ out of $- 125$ .

$\frac{25 \cdot 2}{25 \cdot - 5}$

Step 14.2.2.2

Cancel the common factor.

$\frac{25 \cdot 2}{25 \cdot - 5}$

Step 14.2.2.3

Rewrite the expression.

$\frac{2}{- 5}$

Step 14.3

Move the negative in front of the fraction.

$- \frac{2}{5}$

Step 15

$x = - 5$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 5$ is a local maximum

Step 16

Find the y-value when $x = - 5$ .

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Step 16.1

Replace the variable $x$ with $- 5$ in the expression.

$f (- 5) = (- 5) + \frac{25}{- 5}$

Step 16.2

Simplify the result.

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Step 16.2.1

Divide $25$ by $- 5$ .

$f (- 5) = - 5 - 5$

Step 16.2.2

Subtract $5$ from $- 5$ .

$f (- 5) = - 10$

Step 16.2.3

The final answer is $- 10$ .

$y = - 10$

Step 17

These are the local extrema for $f (x) = x + \frac{25}{x}$ .

$(5, 10)$ is a local minima

$(- 5, - 10)$ is a local maxima

Step 18