Chapter 14.1, Problem 4WE

To determine

To find: The value of $h\left(\frac{1}{2}\right)$ if the function is $h\left(x\right)=6x+1$ and check the function $h$ is a one-to-one or not.

Answer to Problem 4WE

The value of $h\left(\frac{1}{2}\right)$ is 4 and the function $h$ is a one-to-one function.

Explanation of Solution

Calculation:

Consider the given function $h\left(x\right)=6x+1$ .

Substitute $\frac{1}{2}$ for $x$ in the equation.

  $\begin{array}{c}h\left(\frac{1}{2}\right)=6\left(\frac{1}{2}\right)+1\\ =4\end{array}$

So, the value of $h\left(\frac{1}{2}\right)$ is 4.

To check whether a function is one-to-one or not, equation the image with pre image.

Let $h\left(x\right)=y$ .

  $\begin{array}{c}6x+1=y\\ \frac{y-1}{6}=x\end{array}$

From the above equation, it is clear that each image has one pre image value. Therefore, the function $h$ is one-to-one.