Let's consider the function $$f(x)=x^2+1$$.

From its analytical expression we can compute the image of any element $$x$$ in the domain. To do so, it is enough to replace the value of $$x$$ in the expression of the function.

For $$x = 2$$: $$$f(2)=2^2+1=4+1=5$$$

Therefore, $$5$$ is the image of $$2$$ in $$f$$.

We will write $$f (2) = 5$$.

We can calculate also the inverse image or the images of any element $$y$$ of the codomain. To do so, it is enough to replace the value of $$y = f (x)$$ in the expression of the function and to solve $$x$$.

For example, the inverse image of $$y = 10$$ is: $$$\begin{array}{rcl}10&=&x^2+1 \\ x^2&=&9 \\ x&=& \pm 3\end{array}$$$

Therefore, $$3$$ and $$-3$$ are inverse images of $$10$$ for the function $$f$$. We will write: $$$f^{-1}(10)=\{-3, 3\}$$$

Compute the image of $$2$$ and the inverse image of $$11$$ for the function from the previous example $$f(x)=3x^2-1$$.

$$f^{-1}(11)$$: $$$\begin{array}{rcl}11 &=& 3x^2-1 \\ 12 &=& 3x^2 \\ x^2&=& 4\\x&=& \pm2=\{-2,2\}\end{array} \Longrightarrow f^{-1}(11)=\{-2,2\}$$$