High School Math : Understanding Inverse Functions

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Understanding Inverse Functions

Let \displaystyle f(x)=5-(1+x)^3. What is \displaystyle f^{-1}(x)?

Possible Answers:

\displaystyle f^{-1}(x)=(5-x)^{\frac{1}{3}}-1

\displaystyle f^{-1}(x)=(5-x)^{3}-1

Correct answer:

\displaystyle f^{-1}(x)=(5-x)^{\frac{1}{3}}-1

Explanation:

We are asked to find \displaystyle f^{^{-1}} (x), which is the inverse of a function. 

In order to find the inverse, the first thing we want to do is replace f(x) with y. (This usually makes it easier to separate x from its function.).

\displaystyle y=5-(1+x)^3

Next, we will swap x and y.

\displaystyle x=5-(1+y)^3

Then, we will solve for y. The expression that we determine will be equal to \displaystyle f^{^{-1}} (x).

\displaystyle x=5-(1+y)^3

Subtract 5 from both sides.

\displaystyle x-5=-(1+y)^3

Multiply both sides by -1.

\displaystyle -1(x-5)=5-x=(1+y)^3

We need to raise both sides of the equation to the 1/3 power in order to remove the exponent on the right side. 

\displaystyle (5-x)^{\frac{1}{3}}=\left ((1+y)^{3} \right )^{1/3}

We will apply the general property of exponents which states that \displaystyle (a^b)^{c}=a^{b\cdot c}.

\displaystyle (5-x)^{\frac{1}{3}}=\left ((1+y)^{3} \right )^{1/3}=(1+y)^{3\cdot \frac{1}{3}}=(1+y)^1=1+y

Laslty, we will subtract one from both sides.

\displaystyle (5-x)^{\frac{1}{3}}-1=y

The expression equal to y is equal to the inverse of the original function f(x). Thus, we can replace y with \displaystyle f^{-1}(x).

\displaystyle f^{-1}(x)=(5-x)^{\frac{1}{3}}-1

The answer is \displaystyle f^{-1}(x)=(5-x)^{\frac{1}{3}}-1.

Example Question #1 : Understanding Inverse Functions

What is the inverse of \displaystyle y=4x^{2}?

Possible Answers:

\displaystyle y=\frac{\pm \sqrt{x}}{4}

\displaystyle y=-4x^{2}

\displaystyle y=\frac{\pm \sqrt{x}}{2}

\displaystyle y=\frac{x^{2}}{4}

\displaystyle y=\pm \sqrt{x}

Correct answer:

\displaystyle y=\frac{\pm \sqrt{x}}{2}

Explanation:

The inverse of \displaystyle y requires us to interchange \displaystyle x and \displaystyle y and then solve for \displaystyle y.

\displaystyle y=4x^{2} 

\displaystyle x=4y^{2}

Then solve for \displaystyle y:

\displaystyle y=\frac{\pm \sqrt{x}}{2}

Example Question #371 : Algebra Ii

If \displaystyle f(x)=x^{2}+1, what is \displaystyle f^{-1}(x)?

Possible Answers:

\displaystyle f^{-1}(x)=\frac{1}{x^{2}-1}

\displaystyle f^{-1}(x)=x^{2}-1

\displaystyle f^{-1}(x)=\sqrt{x-1}

\displaystyle f^{-1}(x)=\sqrt{x^{2}+1}

\displaystyle f^{-1}(x)=\frac{1}{x^{2}+1}

Correct answer:

\displaystyle f^{-1}(x)=\sqrt{x-1}

Explanation:

To find the inverse of a function, exchange the \displaystyle x and \displaystyle y variables and then solve for \displaystyle y.

\displaystyle x=y^{2}+1

\displaystyle x-1=y^{2}

\displaystyle \sqrt{x-1}=y=f^{-1}(x)

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