Understanding Inverse Functions - Math

Card 0 of 12

Question

Let . What is $f^{-1}(x)$ ?

Answer

We are asked to find $f^{^{-1}} (x)$ , 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.).

Next, we will swap x and y.

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

Subtract 5 from both sides.

Multiply both sides by -1.

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

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

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

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

Laslty, we will subtract one from both sides.

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

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

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

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

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Question

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

Answer

The inverse of requires us to interchange and and then solve for .

$y=4x^{2}$

$x=4y^{2}$

Then solve for :

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

Compare your answer with the correct one above

Question

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

Answer

To find the inverse of a function, exchange the and variables and then solve for .

$x=y^{2}+1$

$x-1=y^{2}$

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

Compare your answer with the correct one above

Question

Let . What is $f^{-1}(x)$ ?

Answer

We are asked to find $f^{^{-1}} (x)$ , 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.).

Next, we will swap x and y.

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

Subtract 5 from both sides.

Multiply both sides by -1.

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

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

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

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

Laslty, we will subtract one from both sides.

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

The inverse of requires us to interchange and and then solve for .

$y=4x^{2}$

$x=4y^{2}$

Then solve for :

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

Compare your answer with the correct one above

Question

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

Answer

To find the inverse of a function, exchange the and variables and then solve for .

$x=y^{2}+1$

$x-1=y^{2}$

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

Compare your answer with the correct one above

Question

Let . What is $f^{-1}(x)$ ?

Answer

We are asked to find $f^{^{-1}} (x)$ , 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.).

Next, we will swap x and y.

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

Subtract 5 from both sides.

Multiply both sides by -1.

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

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

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

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

Laslty, we will subtract one from both sides.

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

The inverse of requires us to interchange and and then solve for .

$y=4x^{2}$

$x=4y^{2}$

Then solve for :

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

Compare your answer with the correct one above

Question

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

Answer

To find the inverse of a function, exchange the and variables and then solve for .

$x=y^{2}+1$

$x-1=y^{2}$

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

Compare your answer with the correct one above

Question

Let . What is $f^{-1}(x)$ ?

Answer

We are asked to find $f^{^{-1}} (x)$ , 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.).

Next, we will swap x and y.

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

Subtract 5 from both sides.

Multiply both sides by -1.

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

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

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

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

Laslty, we will subtract one from both sides.

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

The inverse of requires us to interchange and and then solve for .

$y=4x^{2}$

$x=4y^{2}$

Then solve for :

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

Compare your answer with the correct one above

Question

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

Answer

To find the inverse of a function, exchange the and variables and then solve for .

$x=y^{2}+1$

$x-1=y^{2}$

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

Compare your answer with the correct one above

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