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Calculus Examples

f (x) = 4 x - 3

$f (x) = 4 x - 3$

Step 1

Write

f (x) = 4 x - 3

$f (x) = 4 x - 3$ as an equation.

y = 4 x - 3

$y = 4 x - 3$

Step 2

Interchange the variables.

x = 4 y - 3

$x = 4 y - 3$

Step 3

Solve for

y

$y$ .

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Step 3.1

Rewrite the equation as

4 y - 3 = x

$4 y - 3 = x$ .

4 y - 3 = x

$4 y - 3 = x$

Step 3.2

Add

3

$3$ to both sides of the equation.

4 y = x + 3

$4 y = x + 3$

Step 3.3

Divide each term in

4 y = x + 3

$4 y = x + 3$ by

4

$4$ and simplify.

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Step 3.3.1

Divide each term in

4 y = x + 3

$4 y = x + 3$ by

4

$4$ .

\frac{4 y}{4} = \frac{x}{4} + \frac{3}{4}

$\frac{4 y}{4} = \frac{x}{4} + \frac{3}{4}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of

4

$4$ .

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Step 3.3.2.1.1

Cancel the common factor.

\frac{4 y}{4} = \frac{x}{4} + \frac{3}{4}

$\frac{4 y}{4} = \frac{x}{4} + \frac{3}{4}$

Step 3.3.2.1.2

Divide

y

$y$ by

1

$1$ .

y = \frac{x}{4} + \frac{3}{4}

$y = \frac{x}{4} + \frac{3}{4}$

y = \frac{x}{4} + \frac{3}{4}

$y = \frac{x}{4} + \frac{3}{4}$

y = \frac{x}{4} + \frac{3}{4}

$y = \frac{x}{4} + \frac{3}{4}$

y = \frac{x}{4} + \frac{3}{4}

$y = \frac{x}{4} + \frac{3}{4}$

y = \frac{x}{4} + \frac{3}{4}

$y = \frac{x}{4} + \frac{3}{4}$

Step 4

Replace

y

$y$ with

f^{- 1} (x)

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

f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}

$f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}$

Step 5

Verify if

f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}

$f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}$ is the inverse of

f (x) = 4 x - 3

$f (x) = 4 x - 3$ .

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Step 5.1

To verify the inverse, check if

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

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

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

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

Step 5.2

Evaluate

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

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

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Step 5.2.1

Set up the composite result function.

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

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

Step 5.2.2

Evaluate

f^{- 1} (4 x - 3)

$f^{- 1} (4 x - 3)$ by substituting in the value of

f

$f$ into

f^{- 1}

$f^{- 1}$ .

f^{- 1} (4 x - 3) = \frac{4 x - 3}{4} + \frac{3}{4}

$f^{- 1} (4 x - 3) = \frac{4 x - 3}{4} + \frac{3}{4}$

Step 5.2.3

Combine the numerators over the common denominator.

f^{- 1} (4 x - 3) = \frac{4 x - 3 + 3}{4}

$f^{- 1} (4 x - 3) = \frac{4 x - 3 + 3}{4}$

Step 5.2.4

Combine the opposite terms in

4 x - 3 + 3

$4 x - 3 + 3$ .

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Step 5.2.4.1

Add

- 3

$- 3$ and

3

$3$ .

f^{- 1} (4 x - 3) = \frac{4 x + 0}{4}

$f^{- 1} (4 x - 3) = \frac{4 x + 0}{4}$

Step 5.2.4.2

Add

4 x

$4 x$ and

0

$0$ .

f^{- 1} (4 x - 3) = \frac{4 x}{4}

$f^{- 1} (4 x - 3) = \frac{4 x}{4}$

f^{- 1} (4 x - 3) = \frac{4 x}{4}

$f^{- 1} (4 x - 3) = \frac{4 x}{4}$

Step 5.2.5

Cancel the common factor of

4

$4$ .

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Step 5.2.5.1

Cancel the common factor.

f^{- 1} (4 x - 3) = \frac{4 x}{4}

$f^{- 1} (4 x - 3) = \frac{4 x}{4}$

Step 5.2.5.2

Divide

x

$x$ by

1

$1$ .

f^{- 1} (4 x - 3) = x

$f^{- 1} (4 x - 3) = x$

f^{- 1} (4 x - 3) = x

$f^{- 1} (4 x - 3) = x$

f^{- 1} (4 x - 3) = x

$f^{- 1} (4 x - 3) = x$

Step 5.3

Evaluate

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

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

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Step 5.3.1

Set up the composite result function.

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

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

Step 5.3.2

Evaluate

f (\frac{x}{4} + \frac{3}{4})

$f (\frac{x}{4} + \frac{3}{4})$ by substituting in the value of

f^{- 1}

$f^{- 1}$ into

f

$f$ .

f (\frac{x}{4} + \frac{3}{4}) = 4 (\frac{x}{4} + \frac{3}{4}) - 3

$f (\frac{x}{4} + \frac{3}{4}) = 4 (\frac{x}{4} + \frac{3}{4}) - 3$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

f (\frac{x}{4} + \frac{3}{4}) = 4 (\frac{x}{4}) + 4 (\frac{3}{4}) - 3

$f (\frac{x}{4} + \frac{3}{4}) = 4 (\frac{x}{4}) + 4 (\frac{3}{4}) - 3$

Step 5.3.3.2

Cancel the common factor of

4

$4$ .

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Step 5.3.3.2.1

Cancel the common factor.

f (\frac{x}{4} + \frac{3}{4}) = 4 (\frac{x}{4}) + 4 (\frac{3}{4}) - 3

$f (\frac{x}{4} + \frac{3}{4}) = 4 (\frac{x}{4}) + 4 (\frac{3}{4}) - 3$

Step 5.3.3.2.2

Rewrite the expression.

f (\frac{x}{4} + \frac{3}{4}) = x + 4 (\frac{3}{4}) - 3

$f (\frac{x}{4} + \frac{3}{4}) = x + 4 (\frac{3}{4}) - 3$

f (\frac{x}{4} + \frac{3}{4}) = x + 4 (\frac{3}{4}) - 3

$f (\frac{x}{4} + \frac{3}{4}) = x + 4 (\frac{3}{4}) - 3$

Step 5.3.3.3

Cancel the common factor of

4

$4$ .

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Step 5.3.3.3.1

Cancel the common factor.

f (\frac{x}{4} + \frac{3}{4}) = x + 4 (\frac{3}{4}) - 3

$f (\frac{x}{4} + \frac{3}{4}) = x + 4 (\frac{3}{4}) - 3$

Step 5.3.3.3.2

Rewrite the expression.

f (\frac{x}{4} + \frac{3}{4}) = x + 3 - 3

$f (\frac{x}{4} + \frac{3}{4}) = x + 3 - 3$

f (\frac{x}{4} + \frac{3}{4}) = x + 3 - 3

$f (\frac{x}{4} + \frac{3}{4}) = x + 3 - 3$

f (\frac{x}{4} + \frac{3}{4}) = x + 3 - 3

$f (\frac{x}{4} + \frac{3}{4}) = x + 3 - 3$

Step 5.3.4

Combine the opposite terms in

x + 3 - 3

$x + 3 - 3$ .

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Step 5.3.4.1

Subtract

3

$3$ from

3

$3$ .

f (\frac{x}{4} + \frac{3}{4}) = x + 0

$f (\frac{x}{4} + \frac{3}{4}) = x + 0$

Step 5.3.4.2

Add

x

$x$ and

0

$0$ .

f (\frac{x}{4} + \frac{3}{4}) = x

$f (\frac{x}{4} + \frac{3}{4}) = x$

f (\frac{x}{4} + \frac{3}{4}) = x

$f (\frac{x}{4} + \frac{3}{4}) = x$

f (\frac{x}{4} + \frac{3}{4}) = x

$f (\frac{x}{4} + \frac{3}{4}) = x$

Step 5.4

Since

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

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

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

$f (f^{- 1} (x)) = x$ , then

f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}

$f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}$ is the inverse of

f (x) = 4 x - 3

$f (x) = 4 x - 3$ .

f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}

$f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}$

f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}

$f^{- 1} (x) = \frac{x}{4} + \frac{3}{4}$

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⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$