Calculus Examples

$f (x) = 1 + \frac{5}{x} - \frac{9}{x^{2}}$

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 $1 + \frac{5}{x} - \frac{9}{x^{2}}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\frac{5}{x}] + \frac{d}{d x} [- \frac{9}{x^{2}}]$ .

$f' (x) = \frac{d}{d x} (1) + \frac{d}{d x} (\frac{5}{x}) + \frac{d}{d x} (- \frac{9}{x^{2}})$

Step 1.1.1.2

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

$f' (x) = 0 + \frac{d}{d x} (\frac{5}{x}) + \frac{d}{d x} (- \frac{9}{x^{2}})$

Step 1.1.2

Evaluate $\frac{d}{d x} [\frac{5}{x}]$ .

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Step 1.1.2.1

Since $5$ is constant with respect to $x$ , the derivative of $\frac{5}{x}$ with respect to $x$ is $5 \frac{d}{d x} [\frac{1}{x}]$ .

$f' (x) = 0 + 5 \frac{d}{d x} (\frac{1}{x}) + \frac{d}{d x} (- \frac{9}{x^{2}})$

Step 1.1.2.2

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

$f' (x) = 0 + 5 \frac{d}{d x} (x^{- 1}) + \frac{d}{d x} (- \frac{9}{x^{2}})$

Step 1.1.2.3

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

$f' (x) = 0 + 5 (- x^{- 2}) + \frac{d}{d x} (- \frac{9}{x^{2}})$

Step 1.1.2.4

Multiply $- 1$ by $5$ .

$f' (x) = 0 - 5 x^{- 2} + \frac{d}{d x} (- \frac{9}{x^{2}})$

Step 1.1.3

Evaluate $\frac{d}{d x} [- \frac{9}{x^{2}}]$ .

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Step 1.1.3.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- \frac{9}{x^{2}}$ with respect to $x$ is $- 9 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$f' (x) = 0 - 5 x^{- 2} - 9 \frac{d}{d x} \frac{1}{x^{2}}$

Step 1.1.3.2

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

$f' (x) = 0 - 5 x^{- 2} - 9 \frac{d}{d x} {(x^{2})}^{- 1}$

Step 1.1.3.3

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 1.1.3.3.1

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

$f' (x) = 0 - 5 x^{- 2} - 9 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{2}))$

Step 1.1.3.3.2

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

$f' (x) = 0 - 5 x^{- 2} - 9 (- u^{- 2} \frac{d}{d x} x^{2})$

Step 1.1.3.3.3

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

$f' (x) = 0 - 5 x^{- 2} - 9 (- {(x^{2})}^{- 2} \frac{d}{d x} x^{2})$

Step 1.1.3.4

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

$f' (x) = 0 - 5 x^{- 2} - 9 (- {(x^{2})}^{- 2} (2 x))$

Step 1.1.3.5

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

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Step 1.1.3.5.1

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

$f' (x) = 0 - 5 x^{- 2} - 9 (- x^{2 \cdot - 2} (2 x))$

Step 1.1.3.5.2

Multiply $2$ by $- 2$ .

$f' (x) = 0 - 5 x^{- 2} - 9 (- x^{- 4} (2 x))$

Step 1.1.3.6

Multiply $2$ by $- 1$ .

$f' (x) = 0 - 5 x^{- 2} - 9 (- 2 x^{- 4} x)$

Step 1.1.3.7

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

$f' (x) = 0 - 5 x^{- 2} - 9 (- 2 (x \cdot x^{- 4}))$

Step 1.1.3.8

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

$f' (x) = 0 - 5 x^{- 2} - 9 (- 2 x^{1 - 4})$

Step 1.1.3.9

Subtract $4$ from $1$ .

$f' (x) = 0 - 5 x^{- 2} - 9 (- 2 x^{- 3})$

Step 1.1.3.10

Multiply $- 2$ by $- 9$ .

$f' (x) = 0 - 5 x^{- 2} + 18 x^{- 3}$

Step 1.1.4

Simplify.

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Step 1.1.4.1

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

$f' (x) = 0 - 5 \frac{1}{x^{2}} + 18 x^{- 3}$

Step 1.1.4.2

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

$f' (x) = 0 - 5 \frac{1}{x^{2}} + 18 (\frac{1}{x^{3}})$

Step 1.1.4.3

Combine terms.

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Step 1.1.4.3.1

Combine $- 5$ and $\frac{1}{x^{2}}$ .

$f' (x) = 0 + \frac{- 5}{x^{2}} + 18 (\frac{1}{x^{3}})$

Step 1.1.4.3.2

Move the negative in front of the fraction.

$f' (x) = 0 - \frac{5}{x^{2}} + 18 (\frac{1}{x^{3}})$

Step 1.1.4.3.3

Subtract $\frac{5}{x^{2}}$ from $0$ .

$f' (x) = - \frac{5}{x^{2}} + 18 (\frac{1}{x^{3}})$

Step 1.1.4.3.4

Combine $18$ and $\frac{1}{x^{3}}$ .

$f' (x) = - \frac{5}{x^{2}} + \frac{18}{x^{3}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{5}{x^{2}} + \frac{18}{x^{3}}$ .

$- \frac{5}{x^{2}} + \frac{18}{x^{3}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $- \frac{5}{x^{2}} + \frac{18}{x^{3}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- \frac{5}{x^{2}} + \frac{18}{x^{3}} = 0$

Step 2.2

Find the LCD of the terms in the equation.

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Step 2.2.1

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

$x^{2}, x^{3}, 1$

Step 2.2.2

Since $x^{2}, x^{3}, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1$ then find LCM for the variable part $x^{2}, x^{3}$ .

Step 2.2.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 2.2.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.2.5

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 2.2.6

The factors for $x^{2}$ are $x \cdot x$ , which is $x$ multiplied by each other $2$ times.

$x^{2} = x \cdot x$

$x$ occurs $2$ times.

Step 2.2.7

The factors for $x^{3}$ are $x \cdot x \cdot x$ , which is $x$ multiplied by each other $3$ times.

$x^{3} = x \cdot x \cdot x$

$x$ occurs $3$ times.

Step 2.2.8

The LCM of $x^{2}, x^{3}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x \cdot x \cdot x$

Step 2.2.9

Simplify $x \cdot x \cdot x$ .

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Step 2.2.9.1

Multiply $x$ by $x$ .

$x^{2} \cdot x$

Step 2.2.9.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.2.9.2.1

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

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Step 2.2.9.2.1.1

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

$x^{2} \cdot x^{1}$

Step 2.2.9.2.1.2

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

$x^{2 + 1}$

Step 2.2.9.2.2

Add $2$ and $1$ .

$x^{3}$

Step 2.3

Multiply each term in $- \frac{5}{x^{2}} + \frac{18}{x^{3}} = 0$ by $x^{3}$ to eliminate the fractions.

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Step 2.3.1

Multiply each term in $- \frac{5}{x^{2}} + \frac{18}{x^{3}} = 0$ by $x^{3}$ .

$- \frac{5}{x^{2}} x^{3} + \frac{18}{x^{3}} x^{3} = 0 x^{3}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Simplify each term.

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Step 2.3.2.1.1

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

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Step 2.3.2.1.1.1

Move the leading negative in $- \frac{5}{x^{2}}$ into the numerator.

$\frac{- 5}{x^{2}} x^{3} + \frac{18}{x^{3}} x^{3} = 0 x^{3}$

Step 2.3.2.1.1.2

Factor $x^{2}$ out of $x^{3}$ .

$\frac{- 5}{x^{2}} (x^{2} x) + \frac{18}{x^{3}} x^{3} = 0 x^{3}$

Step 2.3.2.1.1.3

Cancel the common factor.

$\frac{- 5}{x^{2}} (x^{2} x) + \frac{18}{x^{3}} x^{3} = 0 x^{3}$

Step 2.3.2.1.1.4

Rewrite the expression.

$- 5 x + \frac{18}{x^{3}} x^{3} = 0 x^{3}$

Step 2.3.2.1.2

Cancel the common factor of $x^{3}$ .

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Step 2.3.2.1.2.1

Cancel the common factor.

$- 5 x + \frac{18}{x^{3}} x^{3} = 0 x^{3}$

Step 2.3.2.1.2.2

Rewrite the expression.

$- 5 x + 18 = 0 x^{3}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Multiply $0$ by $x^{3}$ .

$- 5 x + 18 = 0$

Step 2.4

Solve the equation.

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Step 2.4.1

Subtract $18$ from both sides of the equation.

$- 5 x = - 18$

Step 2.4.2

Divide each term in $- 5 x = - 18$ by $- 5$ and simplify.

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Step 2.4.2.1

Divide each term in $- 5 x = - 18$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 18}{- 5}$

Step 2.4.2.2

Simplify the left side.

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Step 2.4.2.2.1

Cancel the common factor of $- 5$ .

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Step 2.4.2.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 18}{- 5}$

Step 2.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 18}{- 5}$

Step 2.4.2.3

Simplify the right side.

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Step 2.4.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{18}{5}$

Step 3

The values which make the derivative equal to $0$ are $\frac{18}{5}$ .

$\frac{18}{5}$

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{5}{x^{2}}$ 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 4.3

Set the denominator in $\frac{18}{x^{3}}$ equal to $0$ to find where the expression is undefined.

$x^{3} = 0$

Step 4.4

Solve for $x$ .

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Step 4.4.1

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

$x = \sqrt[3]{0}$

Step 4.4.2

Simplify $\sqrt[3]{0}$ .

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Step 4.4.2.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 4.4.2.2

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

$x = 0$

Step 5

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

$(- \infty, 0) \cup (0, \frac{18}{5}) \cup (\frac{18}{5}, \infty)$

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.

$f' (- 1) = - \frac{5}{{(- 1)}^{2}} + \frac{18}{{(- 1)}^{3}}$

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

$f' (- 1) = - \frac{5}{1} + \frac{18}{{(- 1)}^{3}}$

Step 6.2.1.2

Divide $5$ by $1$ .

$f' (- 1) = - 1 \cdot 5 + \frac{18}{{(- 1)}^{3}}$

Step 6.2.1.3

Multiply $- 1$ by $5$ .

$f' (- 1) = - 5 + \frac{18}{{(- 1)}^{3}}$

Step 6.2.1.4

Raise $- 1$ to the power of $3$ .

$f' (- 1) = - 5 + \frac{18}{- 1}$

Step 6.2.1.5

Divide $18$ by $- 1$ .

$f' (- 1) = - 5 - 18$

Step 6.2.2

Subtract $18$ from $- 5$ .

$f' (- 1) = - 23$

Step 6.2.3

The final answer is $- 23$ .

$- 23$

Step 6.3

At $x = - 1$ the derivative is $- 23$ . 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 $(0, \frac{18}{5})$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1.8$ in the expression.

$f' (1.8) = - \frac{5}{{(1.8)}^{2}} + \frac{18}{{(1.8)}^{3}}$

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 $1.8$ to the power of $2$ .

$f' (1.8) = - \frac{5}{3.24} + \frac{18}{{(1.8)}^{3}}$

Step 7.2.1.2

Divide $5$ by $3.24$ .

$f' (1.8) = - 1 \cdot 1.54320987 + \frac{18}{{(1.8)}^{3}}$

Step 7.2.1.3

Multiply $- 1$ by $1.54320987$ .

$f' (1.8) = - 1.54320987 + \frac{18}{{(1.8)}^{3}}$

Step 7.2.1.4

Raise $1.8$ to the power of $3$ .

$f' (1.8) = - 1.54320987 + \frac{18}{5.832}$

Step 7.2.1.5

Divide $18$ by $5.832$ .

$f' (1.8) = - 1.54320987 + 3.08641975$

Step 7.2.2

Add $- 1.54320987$ and $3.08641975$ .

$f' (1.8) = 1.54320987$

Step 7.2.3

The final answer is $1.54320987$ .

$1.54320987$

Step 7.3

At $x = 1.8$ the derivative is $1.54320987$ . Since this is positive, the function is increasing on $(0, \frac{18}{5})$ .

Increasing on $(0, \frac{18}{5})$ since $f' (x) > 0$

Step 8

Substitute a value from the interval $(\frac{18}{5}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $4.6$ in the expression.

$f' (4.6) = - \frac{5}{{(4.6)}^{2}} + \frac{18}{{(4.6)}^{3}}$

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 $4.6$ to the power of $2$ .

$f' (4.6) = - \frac{5}{21.16} + \frac{18}{{(4.6)}^{3}}$

Step 8.2.1.2

Divide $5$ by $21.16$ .

$f' (4.6) = - 1 \cdot 0.23629489 + \frac{18}{{(4.6)}^{3}}$

Step 8.2.1.3

Multiply $- 1$ by $0.23629489$ .

$f' (4.6) = - 0.23629489 + \frac{18}{{(4.6)}^{3}}$

Step 8.2.1.4

Raise $4.6$ to the power of $3$ .

$f' (4.6) = - 0.23629489 + \frac{18}{97.336}$

Step 8.2.1.5

Divide $18$ by $97.336$ .

$f' (4.6) = - 0.23629489 + 0.18492644$

Step 8.2.2

Add $- 0.23629489$ and $0.18492644$ .

$f' (4.6) = - 0.05136845$

Step 8.2.3

The final answer is $- 0.05136845$ .

$- 0.05136845$

Step 8.3

At $x = 4.6$ the derivative is $- 0.05136845$ . Since this is negative, the function is decreasing on $(\frac{18}{5}, \infty)$ .

Decreasing on $(\frac{18}{5}, \infty)$ since $f' (x) < 0$

Step 9

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

Increasing on: $(0, \frac{18}{5})$

Decreasing on: $(- \infty, 0), (\frac{18}{5}, \infty)$

Step 10