Problems from Concept and equation of a function

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:

$f(-2)=-3$

$f(1)=3$

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

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