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

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Answer 1 · 2015-09-24T16:18:09

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

On simplifying ,
f(x) = ${(x + 2)}^{2}$ $(x+2)^2$ + 8

As 0 $\leq$ $<=$ x $\leq$ $<=$ 5,
2 $\leq$ $<=$ x+2 $\leq$ $<=$ 7
Therefore, 4 $\leq$ $<=$ ${(x + 2)}^{2}$ $(x+2)^2$ $\leq$ $<=$ 49

4+8 $\leq$ $<=$ ${(x + 2)}^{2}$ $(x+2)^2$ $\leq$ $<=$ 49 +8
12 $\leq$ $<=$ ${(x + 2)}^{2}$ $(x+2)^2$ $\leq$ $<=$ 57

How do you find the range of f(x)=x2+4x+12f(x)=x^2+4x+12 with the domain of 0≤x≤50<=x<=5?