Chapter 4.CT, Problem 1CT

(a)

To determine

To find:

The domain and range of the function $f\left(x\right)=\sqrt[3]{2x-7}$

Answer to Problem 1CT

Solution:

The domain and the range of the function is $\left(-\infty , \infty \right)$.

Explanation of Solution

Concept:

Domain:

Domain is all the $x$ values that have an output of $y$ values.

Range:

Range is all the $y$ values or output of the function.

Calculation:

The given function is $f\left(x\right)=y=\sqrt[3]{2x-7}$.

Since the function does not have the undefined points not the domain constraints, the domain of the given function is all real numbers. That is, in the interval notation the domain of the function is $\left(-\infty , \infty \right)$.

The range of the function is $\left(-\infty , \infty \right)$.

Thus, the domain and the range of the function is $\left(-\infty , \infty \right)$.

Final statement:

The domain and the range of the function is $\left(-\infty , \infty \right)$.

(b)

To determine

To explain:

The reason for the existence of $f^{-1}$.

Answer to Problem 1CT

Solution:

The function is an one-to-one function and will have the inverse function $f^{-1}$.

Explanation of Solution

Concept:

For a function to have the inverse the function must be an one-to-one function.

For a function to be one-to-one, $f\left(a\right)=f\left(b\right)$.

Calculation:

The given function is $f\left(x\right)=y=\sqrt[3]{2x-7}$.

$\sqrt[3]{2a-7}=\sqrt[3]{2b-7}$

Taking the above equation to the power of $3$, we get

${\left(\sqrt[3]{2a-7}\right)}^3={\left(\sqrt[3]{2b-7}\right)}^3$

$2a-7=2b-7$

$a=b$

Thus, the function is an one-to-one function and will have the inverse function $f^{-1}$.

Final statement:

The function is an one-to-one function and will have the inverse function $f^{-1}$.

(c)

To determine

To find:

The inverse of the function $f\left(x\right)=\sqrt[3]{2x-7}$

Answer to Problem 1CT

Solution:

The inverse of the function $f\left(x\right)=\sqrt[3]{2x-7}$ is $f^{-1}\left(x\right)=\frac{x^3+7}{2}$.

Explanation of Solution

Concept:

For a one-to-one function $f$ defined by an equation $y=f\left(x\right)$, find the defining equation of the inverse as follows:

(If necessary, replace $f\left(x\right)$ with $y$ first. Any restrictions on $x$ and $y$ should be considered.)

Step 1: Interchange $x$ and $y$.

Step 2: Solve for $y$.

Step 3: Replace $y$ with $f^{-1}\left(x\right)$.

Calculation:

The given function is $f\left(x\right)=y=\sqrt[3]{2x-7}$.

Interchanging the variables $x$ and $y$, we get

$x=\sqrt[3]{2y-7}$

Taking the above equation to the power of $3$, we get

$x^3={\left(\sqrt[3]{2y-7}\right)}^3$

$x^3=2y-7$

$2y=x^3+7$

$y=\frac{x^3+7}{2}$

Thus, the inverse of the function $f\left(x\right)=\sqrt[3]{2x-7}$ is $f^{-1}\left(x\right)=\frac{x^3+7}{2}$.

Final statement:

The inverse of the function $f\left(x\right)=\sqrt[3]{2x-7}$ is $f^{-1}\left(x\right)=\frac{x^3+7}{2}$.

(d)

To determine

To find:

The domain and range of the inverse function $f^{-1}\left(x\right)=\frac{x^3+7}{2}$

Answer to Problem 1CT

Solution:

The domain and the range of the inverse function is $\left(-\infty , \infty \right)$.

Explanation of Solution

Concept:

Domain:

Domain is all the $x$ values that have an output of $y$ values.

Range:

Range is all the $y$ values or output of the function.

Calculation:

The inverse function is $f^{-1}\left(x\right)=\frac{x^3+7}{2}$.

Since the function does not have the undefined points not the domain constraints, the domain of the given function is all real numbers. That is, in the interval notation the domain of the function is $\left(-\infty , \infty \right)$.

The range of the function is $\left(-\infty , \infty \right)$.

Thus, the domain and the range of the function is $\left(-\infty , \infty \right)$.

Final statement:

The domain and the range of the function is $\left(-\infty , \infty \right)$.

(e)

To determine

To plot:

The graph of the functions $f$ and $f^{-1}$.

Answer to Problem 1CT

Solution:

The graphs of the function $f\left(x\right)=\sqrt[3]{2x-7}$ and the inverse function $f^{-1}\left(x\right)=\frac{x^3+7}{2}$ are plotted.

Explanation of Solution

Concept:

For a one-to-one function $f$ defined by an equation $y=f\left(x\right)$, find the defining equation of the inverse as follows:

(If necessary, replace $f\left(x\right)$ with $y$ first. Any restrictions on $x$ and $y$ should be considered.)

Step 1: Interchange $x$ and $y$.

Step 2: Solve for $y$.

Step 3: Replace $y$ with $f^{-1}\left(x\right)$.

Calculation:

The graphs of the function $f\left(x\right)=\sqrt[3]{2x-7}$ and the inverse function $f^{-1}\left(x\right)=\frac{x^3+7}{2}$ are plotted as shown.

Final statement:

The graphs of the function $f\left(x\right)=\sqrt[3]{2x-7}$ and the inverse function $f^{-1}\left(x\right)=\frac{x^3+7}{2}$ are plotted.