Question

If $h\left(x\right)=6x+1$, find $h\left(\frac{1}{2}\right)$.Is $h$ a one-to-one function?

Short answer

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

Step-by-step solution

Step 1. Given Information.

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

Step 2. Calculation.

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

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

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

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

Step 3. Conclusion.

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