Calculus Examples

$f (x) = \frac{1}{5} x + 7$

Step 1

Write $f (x) = \frac{1}{5} x + 7$ as an equation.

$y = \frac{1}{5} x + 7$

Step 2

Interchange the variables.

$x = \frac{1}{5} y + 7$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{1}{5} y + 7 = x$ .

$\frac{1}{5} y + 7 = x$

Step 3.2

Combine $\frac{1}{5}$ and $y$ .

$\frac{y}{5} + 7 = x$

Step 3.3

Subtract $7$ from both sides of the equation.

$\frac{y}{5} = x - 7$

Step 3.4

Multiply both sides of the equation by $5$ .

$5 \frac{y}{5} = 5 (x - 7)$

Step 3.5

Simplify both sides of the equation.

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Step 3.5.1

Simplify the left side.

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Step 3.5.1.1

Cancel the common factor of $5$ .

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Step 3.5.1.1.1

Cancel the common factor.

$5 \frac{y}{5} = 5 (x - 7)$

Step 3.5.1.1.2

Rewrite the expression.

$y = 5 (x - 7)$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify $5 (x - 7)$ .

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Step 3.5.2.1.1

Apply the distributive property.

$y = 5 x + 5 \cdot - 7$

Step 3.5.2.1.2

Multiply $5$ by $- 7$ .

$y = 5 x - 35$

Step 4

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = 5 x - 35$

Step 5

Verify if $f^{- 1} (x) = 5 x - 35$ is the inverse of $f (x) = \frac{1}{5} x + 7$ .

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Step 5.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 5.2

Evaluate $f^{- 1} (f (x))$ .

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate $f^{- 1} (\frac{1}{5} x + 7)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{5} x + 7) = 5 (\frac{1}{5} x + 7) - 35$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Combine $\frac{1}{5}$ and $x$ .

$f^{- 1} (\frac{1}{5} x + 7) = 5 (\frac{x}{5} + 7) - 35$

Step 5.2.3.2

Apply the distributive property.

$f^{- 1} (\frac{1}{5} x + 7) = 5 (\frac{x}{5}) + 5 \cdot 7 - 35$

Step 5.2.3.3

Cancel the common factor of $5$ .

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Step 5.2.3.3.1

Cancel the common factor.

$f^{- 1} (\frac{1}{5} x + 7) = 5 (\frac{x}{5}) + 5 \cdot 7 - 35$

Step 5.2.3.3.2

Rewrite the expression.

$f^{- 1} (\frac{1}{5} x + 7) = x + 5 \cdot 7 - 35$

Step 5.2.3.4

Multiply $5$ by $7$ .

$f^{- 1} (\frac{1}{5} x + 7) = x + 35 - 35$

Step 5.2.4

Combine the opposite terms in $x + 35 - 35$ .

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Step 5.2.4.1

Subtract $35$ from $35$ .

$f^{- 1} (\frac{1}{5} x + 7) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\frac{1}{5} x + 7) = x$

Step 5.3

Evaluate $f (f^{- 1} (x))$ .

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $f (5 x - 35)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (5 x - 35) = \frac{1}{5} \cdot (5 x - 35) + 7$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

$f (5 x - 35) = \frac{1}{5} \cdot (5 x) + \frac{1}{5} \cdot - 35 + 7$

Step 5.3.3.2

Cancel the common factor of $5$ .

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Step 5.3.3.2.1

Factor $5$ out of $5 x$ .

$f (5 x - 35) = \frac{1}{5} \cdot (5 (x)) + \frac{1}{5} \cdot - 35 + 7$

Step 5.3.3.2.2

Cancel the common factor.

$f (5 x - 35) = \frac{1}{5} \cdot (5 x) + \frac{1}{5} \cdot - 35 + 7$

Step 5.3.3.2.3

Rewrite the expression.

$f (5 x - 35) = x + \frac{1}{5} \cdot - 35 + 7$

Step 5.3.3.3

Cancel the common factor of $5$ .

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Step 5.3.3.3.1

Factor $5$ out of $- 35$ .

$f (5 x - 35) = x + \frac{1}{5} \cdot (5 (- 7)) + 7$

Step 5.3.3.3.2

Cancel the common factor.

$f (5 x - 35) = x + \frac{1}{5} \cdot (5 \cdot - 7) + 7$

Step 5.3.3.3.3

Rewrite the expression.

$f (5 x - 35) = x - 7 + 7$

Step 5.3.4

Combine the opposite terms in $x - 7 + 7$ .

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Step 5.3.4.1

Add $- 7$ and $7$ .

$f (5 x - 35) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (5 x - 35) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 5 x - 35$ is the inverse of $f (x) = \frac{1}{5} x + 7$ .

$f^{- 1} (x) = 5 x - 35$