Question

what is the inverse of the function f(x) = 2x + 1?
○ h(x)=\frac{1}{2}x-\frac{1}{2}
○ h(x)=\frac{1}{2}x+\frac{1}{2}
○ h(x)=\frac{1}{2}x - 2
○ h(x)=\frac{1}{2}x + 2

Explanation:

Step1: Let $y = f(x)$

$y = 2x+17$

Step2: Solve for $x$ in terms of $y$

$y-17 = 2x$, then $x=\frac{y - 17}{2}=\frac{1}{2}y-\frac{17}{2}$

Step3: Replace $x$ with $h(x)$ and $y$ with $x$

$h(x)=\frac{1}{2}x-\frac{17}{2}$

However, if we assume there is a mistake in the problem - setup and we just want to follow the process of finding the inverse of $y = 2x + 1$:

Step1: Let $y = 2x+1$

Step2: Solve for $x$

$y - 1=2x$, so $x=\frac{y - 1}{2}=\frac{1}{2}y-\frac{1}{2}$

Step3: Replace $x$ with $h(x)$ and $y$ with $x$

$h(x)=\frac{1}{2}x-\frac{1}{2}$

Answer:

$h(x)=\frac{1}{2}x-\frac{1}{2}$ (assuming the function was meant to be $f(x)=2x + 1$)