Question

How do you find the range of `f(x)=x^2+4x+12` with the domain of `0<=x<=5`?

Answer 1

The answer is 12 `<=` f(x) `<=`57

Explanation:

On simplifying ,
f(x) = `(x+2)^2` + 8

As 0`<=`x`<=`5,
2`<=`x+2`<=`7
Therefore, 4`<=` `(x+2)^2` `<=`49

4+8 `<=` `(x+2)^2` `<=`49 +8
12 `<=` `(x+2)^2` `<=`57