Maximum and minimum value of equations #2

If we take the derivative of $P$ , we get $P’=\frac{2(x^{2}+1)}{(x^{2}+x+1)^{2}}$ .

Setting it equal to $0$ , we get $\frac{2(x^{2}-1)}{(x^{2}+x+1)^{2}}=0$

$\implies 2(x^{2}-1)=0$

$\implies x^{2}-1=0$

$\implies x^{2}=1$

$\implies x= \pm 1$

$P(1)=\frac{1^{2}-1+1}{1^{2}+1+1}=\frac{1}{3}$

$P(-1)=\frac{(-1)^{2}-(-1)+1}{(-1)^{2}+(-1)+1}=3$

However, we must check the values of $\displaystyle{\lim_{x \to \infty}} P(x)$ and $\displaystyle{\lim_{x \to -\infty}} P(x)$

Luckily, both of them are equal to $1$ (L’hospital’s rule).

Therefore, the sum of the minimum and maximum of $P$ is $3+\frac{1}{3} = \frac{10}{3}$

This means that $a+b=10+3=\fbox{13}$