Consider the function $$f(x) = 2x+1$$ and calculate $$f(-2), f (1), f^{-1}(1)$$.

See development and solution

### Development:

To calculate the images of $$-2$$ and $$1$$ it is enough to substitute in the function:

$$f (-2) = 2 \cdot (-2) +1 =-4 + 1 =-3 $$

$$f (1) = 2 \cdot 1 +1 = 2 + 1 = 3 $$

To calculate the antiimage at point $$1$$, that is to say $$f^{-1}(1)$$, we must equal the expression of the function to the number of the antiimage that we want to calculate, and solve the resultant equation:

$$f^{-1}(1): \ 1= 2x + 1$$

$$2x = 1 - 1 = 0$$

$$x = 0$$

Therefore, $$f^{-1}(1)=0$$

### Solution:

$$f(-2)=-3 $$

$$f(1)=3 $$

$$f^{-1}(1)=0 $$