Inverse Functions Practice

Let's start by replacing $f(x)$ with $y$.

$y=-6x-7$

If a function contains the point $(a,b)$, the inverse of that function contains the point $(b,a)$.

So if we swap the position of $x$ and $y$ in the equation, we get the inverse relationship.

In this case, the function is $y=-6x-7$, so the inverse relationship is $x=-6y-7$.

Solving this equation for $y$ will give us an expression for $f^{-1}(x)$.

$\begin{aligned} \qquad x&=-6y-7\\\\x+7&=-6y\\\\-\dfrac{1}{6}(x+7)&=y\end{aligned}$

The inverse of the function is $f^{-1}(x)=-\dfrac{1}{6}(x+7)$.

Let's start by replacing $f(x)$ with$y$.

$y=-\dfrac{1}{2}(x+3)$

If a function contains the point $(a,b)$, the inverse of that function contains the point $(b,a)$.

So if we swap the position of $x$ and $y$ in the equation, we get the inverse relationship.

In this case, the function is $y=-\dfrac{1}{2}(x+3)$, so the inverse relationship is $x=-\dfrac{1}{2}(y+3)$.

Solving this equation for $y$ will give us an expression for $f^{-1}(x)$.

$\begin{aligned} \qquad x&=-\dfrac{1}{2}(y+3)\\\\-2x&=y+3\\\\-2x-3&=y\end{aligned}$

The inverse of the function is $f^{-1}(x)=-2x-3$.

Let's start by replacing $f(x)$ with $y$.

$y=8x+1$

If a function contains the point $(a,b)$, the inverse of that function contains the point $(b,a)$.

So if we swap the position of $x$ and $y$ in the equation, we get the inverse relationship.

In this case, the function is $y=8x+1$, so the inverse relationship is $x=8y+1$.

Solving this equation for $y$ will give us an expression for $f^{-1}(x)$.

$\begin{aligned} \qquad x&=8y+1\\\\x-1&=8y\\\\\dfrac{x-1}{8}&=y\end{aligned}$

The inverse of the function is  $f^{-1}(x)=\dfrac{x-1}{8}$.

Let's start by replacing $g(x)$ with $y$.

$y=5(x-2)$

If a function contains the point $(a,b)$, the inverse of that function contains the point $(b,a)$.

So if we swap the position of $x$ and $y$ in the equation, we get the inverse relationship.

In this case, the function is $y=5(x-2)$, so the inverse relationship is $x=5(y-2)$.

Solving this equation for $y$ will give us an expression for $g^{-1}(x)$.

$\begin{aligned} \qquad x&=5(y-2)\\\\\dfrac{1}{5}x&=y-2\\\\\dfrac{1}{5}x+2&=y\end{aligned}$

The inverse of the function is $g^{-1}(x)=\dfrac{1}{5}x+2$.