Precalculus Examples

$w = 50 + 2.3 (h - 60)$

Step 1

Interchange the variables.

$h = 50 + 2.3 (y - 60)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $50 + 2.3 (y - 60) = h$ .

$50 + 2.3 (y - 60) = h$

Step 2.2

Simplify $50 + 2.3 (y - 60)$ .

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Apply the distributive property.

$50 + 2.3 y + 2.3 \cdot - 60 = h$

Step 2.2.1.2

Multiply $2.3$ by $- 60$ .

$50 + 2.3 y - 138 = h$

Step 2.2.2

Subtract $138$ from $50$ .

$2.3 y - 88 = h$

Step 2.3

Add $88$ to both sides of the equation.

$2.3 y = h + 88$

Step 2.4

Divide each term in $2.3 y = h + 88$ by $2.3$ and simplify.

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Step 2.4.1

Divide each term in $2.3 y = h + 88$ by $2.3$ .

$\frac{2.3 y}{2.3} = \frac{h}{2.3} + \frac{88}{2.3}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Cancel the common factor of $2.3$ .

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Step 2.4.2.1.1

Cancel the common factor.

$\frac{2.3 y}{2.3} = \frac{h}{2.3} + \frac{88}{2.3}$

Step 2.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{h}{2.3} + \frac{88}{2.3}$

Step 2.4.3

Simplify the right side.

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Step 2.4.3.1

Simplify each term.

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Step 2.4.3.1.1

Multiply by $1$ .

$y = \frac{1 h}{2.3} + \frac{88}{2.3}$

Step 2.4.3.1.2

Factor $2.3$ out of $2.3$ .

$y = \frac{1 h}{2.3 (1)} + \frac{88}{2.3}$

Step 2.4.3.1.3

Separate fractions.

$y = \frac{1}{2.3} \cdot \frac{h}{1} + \frac{88}{2.3}$

Step 2.4.3.1.4

Divide $1$ by $2.3$ .

$y = 0.4347826 \frac{h}{1} + \frac{88}{2.3}$

Step 2.4.3.1.5

Divide $h$ by $1$ .

$y = 0.4347826 h + \frac{88}{2.3}$

Step 2.4.3.1.6

Divide $88$ by $2.3$ .

$y = 0.4347826 h + 38.26086956$

Step 3

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

$f^{- 1} (h) = 0.4347826 h + 38.26086956$

Step 4

Verify if $f^{- 1} (h) = 0.4347826 h + 38.26086956$ is the inverse of $f (h) = 50 + 2.3 (h - 60)$ .

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate $f^{- 1} (50 + 2.3 (h - 60))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (50 + 2.3 (h - 60)) = 0.4347826 (50 + 2.3 (h - 60)) + 38.26086956$

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Simplify each term.

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Step 4.2.3.1.1

Apply the distributive property.

$f^{- 1} (50 + 2.3 (h - 60)) = 0.4347826 (50 + 2.3 h + 2.3 \cdot - 60) + 38.26086956$

Step 4.2.3.1.2

Multiply $2.3$ by $- 60$ .

$f^{- 1} (50 + 2.3 (h - 60)) = 0.4347826 (50 + 2.3 h - 138) + 38.26086956$

Step 4.2.3.2

Subtract $138$ from $50$ .

$f^{- 1} (50 + 2.3 (h - 60)) = 0.4347826 (2.3 h - 88) + 38.26086956$

Step 4.2.3.3

Apply the distributive property.

$f^{- 1} (50 + 2.3 (h - 60)) = 0.4347826 (2.3 h) + 0.4347826 \cdot - 88 + 38.26086956$

Step 4.2.3.4

Multiply $2.3$ by $0.4347826$ .

$f^{- 1} (50 + 2.3 (h - 60)) = 1 h + 0.4347826 \cdot - 88 + 38.26086956$

Step 4.2.3.5

Multiply $0.4347826$ by $- 88$ .

$f^{- 1} (50 + 2.3 (h - 60)) = 1 h - 38.26086956 + 38.26086956$

Step 4.2.4

Add $- 38.26086956$ and $38.26086956$ .

$f^{- 1} (50 + 2.3 (h - 60)) = 1 h 0$

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate $f (0.4347826 h + 38.26086956)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (0.4347826 h + 38.26086956) = 50 + 2.3 ((0.4347826 h + 38.26086956) - 60)$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Subtract $60$ from $38.26086956$ .

$f (0.4347826 h + 38.26086956) = 50 + 2.3 (0.4347826 h - 21.73913043)$

Step 4.3.3.2

Apply the distributive property.

$f (0.4347826 h + 38.26086956) = 50 + 2.3 (0.4347826 h) + 2.3 \cdot - 21.73913043$

Step 4.3.3.3

Multiply $0.4347826$ by $2.3$ .

$f (0.4347826 h + 38.26086956) = 50 + 1 h + 2.3 \cdot - 21.73913043$

Step 4.3.3.4

Multiply $2.3$ by $- 21.73913043$ .

$f (0.4347826 h + 38.26086956) = 50 + 1 h - 50$

Step 4.3.4

Simplify by subtracting numbers.

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Step 4.3.4.1

Subtract $50$ from $50$ .

$f (0.4347826 h + 38.26086956) = 1 h + 0$

Step 4.3.4.2

Add $1 h$ and $0$ .

$f (0.4347826 h + 38.26086956) = 1 h$

Step 4.4

Since $f^{- 1} (f (h)) = h$ and $f (f^{- 1} (h)) = h$ , then $f^{- 1} (h) = 0.4347826 h + 38.26086956$ is the inverse of $f (h) = 50 + 2.3 (h - 60)$ .

$f^{- 1} (h) = 0.4347826 h + 38.26086956$