Let $P(x)=(m^2+4m+5)x^2-4x+7,m\in R$.If $3\leq x\leq 5$,then find the minimum of the minimum value of $P(x).$

Question

Let $P(x)=(m^2+4m+5)x^2-4x+7,m\in R$.If $3\leq x\leq 5$,then find the minimum of the minimum value of $P(x).$

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The minimum value of $P(x)=(m^2+4m+5)x^2-4x+7$ occurs at $x=\frac{2}{m^2+4m+5}$
So the minimum value of the $P(x)$ is $\frac{-4}{m^2+4m+5}+7$

But i dont know how to find the minimum value of $\frac{-4}{m^2+4m+5}+7$
Please help me.Thanks.

Answer 1

$$\frac{-4}{m^2+4m+5}+7=\frac{-4}{(m+2)^2+1}+7\ge\frac{-4}{1}+7=3$$ The equality happens when $m=-2$. But when $m=-2$, $x=\frac{2}{m^2+4m+5}=2$, that is out of the range $3\le x \le 5$.

Therefore the minimum of $P(x)$ happens at either $x=3$ or $x=5$.

i) $x=3$ $$P(3)=9((m+2)^2+1)\cdot9-4\cdot3+7=9(m+2)^2+4\le4$$

ii) $x=5$ $$P(5)=25((m+2)^2+1)-4\cdot5+7=25(m+2)^2+12\le 12$$

So, the minimum is $4$.