Calculus Examples

$f (x) = 19 x + \frac{1}{x}$

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

By the Sum Rule, the derivative of $19 x + \frac{1}{x}$ with respect to $x$ is $\frac{d}{d x} [19 x] + \frac{d}{d x} [\frac{1}{x}]$ .

$f' (x) = \frac{d}{d x} (19 x) + \frac{d}{d x} (\frac{1}{x})$

Step 1.1.2

Evaluate $\frac{d}{d x} [19 x]$ .

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Step 1.1.2.1

Since $19$ is constant with respect to $x$ , the derivative of $19 x$ with respect to $x$ is $19 \frac{d}{d x} [x]$ .

$f' (x) = 19 \frac{d}{d x} (x) + \frac{d}{d x} (\frac{1}{x})$

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 $n = 1$ .

$f' (x) = 19 \cdot 1 + \frac{d}{d x} (\frac{1}{x})$

Step 1.1.2.3

Multiply $19$ by $1$ .

$f' (x) = 19 + \frac{d}{d x} (\frac{1}{x})$

Step 1.1.3

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

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Step 1.1.3.1

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

$f' (x) = 19 + \frac{d}{d x} (x^{- 1})$

Step 1.1.3.2

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

$f' (x) = 19 - x^{- 2}$

Step 1.1.4

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

$f' (x) = 19 - \frac{1}{x^{2}}$

Step 1.1.5

Reorder terms.

$f' (x) = - \frac{1}{x^{2}} + 19$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{1}{x^{2}} + 19$ .

$- \frac{1}{x^{2}} + 19$

Step 2

Set the first derivative equal to $0$ then solve the equation $- \frac{1}{x^{2}} + 19 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- \frac{1}{x^{2}} + 19 = 0$

Step 2.2

Subtract $19$ from both sides of the equation.

$- \frac{1}{x^{2}} = - 19$

Step 2.3

Find the LCD of the terms in the equation.

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Step 2.3.1

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 2.3.2

The LCM of one and any expression is the expression.

$x^{2}$

Step 2.4

Multiply each term in $- \frac{1}{x^{2}} = - 19$ by $x^{2}$ to eliminate the fractions.

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Step 2.4.1

Multiply each term in $- \frac{1}{x^{2}} = - 19$ by $x^{2}$ .

$- \frac{1}{x^{2}} x^{2} = - 19 x^{2}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

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

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Step 2.4.2.1.1

Move the leading negative in $- \frac{1}{x^{2}}$ into the numerator.

$\frac{- 1}{x^{2}} x^{2} = - 19 x^{2}$

Step 2.4.2.1.2

Cancel the common factor.

$\frac{- 1}{x^{2}} x^{2} = - 19 x^{2}$

Step 2.4.2.1.3

Rewrite the expression.

$- 1 = - 19 x^{2}$

Step 2.5

Solve the equation.

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Step 2.5.1

Rewrite the equation as $- 19 x^{2} = - 1$ .

$- 19 x^{2} = - 1$

Step 2.5.2

Divide each term in $- 19 x^{2} = - 1$ by $- 19$ and simplify.

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Step 2.5.2.1

Divide each term in $- 19 x^{2} = - 1$ by $- 19$ .

$\frac{- 19 x^{2}}{- 19} = \frac{- 1}{- 19}$

Step 2.5.2.2

Simplify the left side.

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Step 2.5.2.2.1

Cancel the common factor of $- 19$ .

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Step 2.5.2.2.1.1

Cancel the common factor.

$\frac{- 19 x^{2}}{- 19} = \frac{- 1}{- 19}$

Step 2.5.2.2.1.2

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

$x^{2} = \frac{- 1}{- 19}$

Step 2.5.2.3

Simplify the right side.

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Step 2.5.2.3.1

Dividing two negative values results in a positive value.

$x^{2} = \frac{1}{19}$

Step 2.5.3

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

$x = \pm \sqrt{\frac{1}{19}}$

Step 2.5.4

Simplify $\pm \sqrt{\frac{1}{19}}$ .

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Step 2.5.4.1

Rewrite $\sqrt{\frac{1}{19}}$ as $\frac{\sqrt{1}}{\sqrt{19}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{19}}$

Step 2.5.4.2

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{19}}$

Step 2.5.4.3

Multiply $\frac{1}{\sqrt{19}}$ by $\frac{\sqrt{19}}{\sqrt{19}}$ .

$x = \pm \frac{1}{\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}$

Step 2.5.4.4

Combine and simplify the denominator.

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Step 2.5.4.4.1

Multiply $\frac{1}{\sqrt{19}}$ by $\frac{\sqrt{19}}{\sqrt{19}}$ .

$x = \pm \frac{\sqrt{19}}{\sqrt{19} \sqrt{19}}$

Step 2.5.4.4.2

Raise $\sqrt{19}$ to the power of $1$ .

$x = \pm \frac{\sqrt{19}}{{\sqrt{19}}^{1} \sqrt{19}}$

Step 2.5.4.4.3

Raise $\sqrt{19}$ to the power of $1$ .

$x = \pm \frac{\sqrt{19}}{{\sqrt{19}}^{1} {\sqrt{19}}^{1}}$

Step 2.5.4.4.4

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

$x = \pm \frac{\sqrt{19}}{{\sqrt{19}}^{1 + 1}}$

Step 2.5.4.4.5

Add $1$ and $1$ .

$x = \pm \frac{\sqrt{19}}{{\sqrt{19}}^{2}}$

Step 2.5.4.4.6

Rewrite ${\sqrt{19}}^{2}$ as $19$ .

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Step 2.5.4.4.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{19}$ as $19^{\frac{1}{2}}$ .

$x = \pm \frac{\sqrt{19}}{{(19^{\frac{1}{2}})}^{2}}$

Step 2.5.4.4.6.2

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

$x = \pm \frac{\sqrt{19}}{19^{\frac{1}{2} \cdot 2}}$

Step 2.5.4.4.6.3

Combine $\frac{1}{2}$ and $2$ .

$x = \pm \frac{\sqrt{19}}{19^{\frac{2}{2}}}$

Step 2.5.4.4.6.4

Cancel the common factor of $2$ .

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Step 2.5.4.4.6.4.1

Cancel the common factor.

$x = \pm \frac{\sqrt{19}}{19^{\frac{2}{2}}}$

Step 2.5.4.4.6.4.2

Rewrite the expression.

$x = \pm \frac{\sqrt{19}}{19^{1}}$

Step 2.5.4.4.6.5

Evaluate the exponent.

$x = \pm \frac{\sqrt{19}}{19}$

Step 2.5.5

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

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Step 2.5.5.1

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

$x = \frac{\sqrt{19}}{19}$

Step 2.5.5.2

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

$x = - \frac{\sqrt{19}}{19}$

Step 2.5.5.3

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

$x = \frac{\sqrt{19}}{19}, - \frac{\sqrt{19}}{19}$

Step 3

The values which make the derivative equal to $0$ are $\frac{\sqrt{19}}{19}, - \frac{\sqrt{19}}{19}$ .

$\frac{\sqrt{19}}{19}, - \frac{\sqrt{19}}{19}$

Step 4

Find where the derivative is undefined.

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Step 4.1

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

$x^{2} = 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

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

$x = \pm \sqrt{0}$

Step 4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 4.2.2.1

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

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

Step 4.2.2.2

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

$x = \pm 0$

Step 4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 5

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - \frac{\sqrt{19}}{19}) \cup (- \frac{\sqrt{19}}{19}, 0) \cup (0, \frac{\sqrt{19}}{19}) \cup (\frac{\sqrt{19}}{19}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - \frac{\sqrt{19}}{19})$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1.22941573$ in the expression.

$f' (- 1.22941573) = - \frac{1}{{(- 1.22941573)}^{2}} + 19$

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

Raise $- 1.22941573$ to the power of $2$ .

$f' (- 1.22941573) = - \frac{1}{1.51146303} + 19$

Step 6.2.1.2

Divide $1$ by $1.51146303$ .

$f' (- 1.22941573) = - 1 \cdot 0.66161062 + 19$

Step 6.2.1.3

Multiply $- 1$ by $0.66161062$ .

$f' (- 1.22941573) = - 0.66161062 + 19$

Step 6.2.2

Add $- 0.66161062$ and $19$ .

$f' (- 1.22941573) = 18.33838937$

Step 6.2.3

The final answer is $18.33838937$ .

$18.33838937$

Step 6.3

At $x = - 1.22941573$ the derivative is $18.33838937$ . Since this is positive, the function is increasing on $(- \infty, - \frac{\sqrt{19}}{19})$ .

Increasing on $(- \infty, - \frac{\sqrt{19}}{19})$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(- 0.22941573, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- 0.11470786$ in the expression.

$f' (- 0.11470786) = - \frac{1}{{(- 0.11470786)}^{2}} + 19$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $- 0.11470786$ to the power of $2$ .

$f' (- 0.11470786) = - \frac{1}{0.01315789} + 19$

Step 7.2.1.2

Divide $1$ by $0.01315789$ .

$f' (- 0.11470786) = - 1 \cdot 76.00000256 + 19$

Step 7.2.1.3

Multiply $- 1$ by $76.00000256$ .

$f' (- 0.11470786) = - 76.00000256 + 19$

Step 7.2.2

Add $- 76.00000256$ and $19$ .

$f' (- 0.11470786) = - 57.00000256$

Step 7.2.3

The final answer is $- 57.00000256$ .

$- 57.00000256$

Step 7.3

At $x = - 0.11470786$ the derivative is $- 57.00000256$ . Since this is negative, the function is decreasing on $(- 0.22941573, 0)$ .

Decreasing on $(- \frac{\sqrt{19}}{19}, 0)$ since $f' (x) < 0$

Step 8

Substitute a value from the interval $(0, \frac{\sqrt{19}}{19})$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $0.11470786$ in the expression.

$f' (0.11470786) = - \frac{1}{{(0.11470786)}^{2}} + 19$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise $0.11470786$ to the power of $2$ .

$f' (0.11470786) = - \frac{1}{0.01315789} + 19$

Step 8.2.1.2

Divide $1$ by $0.01315789$ .

$f' (0.11470786) = - 1 \cdot 76.00000256 + 19$

Step 8.2.1.3

Multiply $- 1$ by $76.00000256$ .

$f' (0.11470786) = - 76.00000256 + 19$

Step 8.2.2

Add $- 76.00000256$ and $19$ .

$f' (0.11470786) = - 57.00000256$

Step 8.2.3

The final answer is $- 57.00000256$ .

$- 57.00000256$

Step 8.3

At $x = 0.11470786$ the derivative is $- 57.00000256$ . Since this is negative, the function is decreasing on $(0, \frac{\sqrt{19}}{19})$ .

Decreasing on $(0, \frac{\sqrt{19}}{19})$ since $f' (x) < 0$

Step 9

Substitute a value from the interval $(\frac{\sqrt{19}}{19}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 9.1

Replace the variable $x$ with $1.22941573$ in the expression.

$f' (1.22941573) = - \frac{1}{{(1.22941573)}^{2}} + 19$

Step 9.2

Simplify the result.

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Step 9.2.1

Simplify each term.

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Step 9.2.1.1

Raise $1.22941573$ to the power of $2$ .

$f' (1.22941573) = - \frac{1}{1.51146303} + 19$

Step 9.2.1.2

Divide $1$ by $1.51146303$ .

$f' (1.22941573) = - 1 \cdot 0.66161062 + 19$

Step 9.2.1.3

Multiply $- 1$ by $0.66161062$ .

$f' (1.22941573) = - 0.66161062 + 19$

Step 9.2.2

Add $- 0.66161062$ and $19$ .

$f' (1.22941573) = 18.33838937$

Step 9.2.3

The final answer is $18.33838937$ .

$18.33838937$

Step 9.3

At $x = 1.22941573$ the derivative is $18.33838937$ . Since this is positive, the function is increasing on $(\frac{\sqrt{19}}{19}, \infty)$ .

Increasing on $(\frac{\sqrt{19}}{19}, \infty)$ since $f' (x) > 0$

Step 10

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

Increasing on: $(- \infty, - \frac{\sqrt{19}}{19}), (\frac{\sqrt{19}}{19}, \infty)$

Decreasing on: $(- \frac{\sqrt{19}}{19}, 0), (0, \frac{\sqrt{19}}{19})$

Step 11