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

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Answer 1 · 2015-09-24T16:18:09

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

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

How do you find the range of f(x)=x^2+4x+12 f(x)=x^2+4x+12 with the domain of 0<=x<=50<=x<=5?