# What are the critical values, if any, of f(x)= (x+1) /(x^2 + x + 1)?

Nov 10, 2016

The critical values will occur when the derivative is $0$ or undefined.

$f ' \left(x\right) = \frac{1 \left({x}^{2} + x + 1\right) - \left(x + 1\right) \left(2 x + 1\right)}{{x}^{2} + x + 1} ^ 2$

$f ' \left(x\right) = \frac{{x}^{2} + x + 1 - \left(2 {x}^{2} + 3 x + 1\right)}{{x}^{2} + x + 1} ^ 2$

$f ' \left(x\right) = \frac{- {x}^{2} - 2 x}{{x}^{2} + x + 1} ^ 2$

Set the derivative to $0$ and solve. Also, find the vertical asymptotes (where the function is undefined).

$0 = \frac{- {x}^{2} - 2 x}{{x}^{2} + x + 1} ^ 2$

$0 = - {x}^{2} - 2 x$

$0 = - x \left(x + 2\right)$

$x = 0 \mathmr{and} - 2$

For V.A:

${\left({x}^{2} + x + 1\right)}^{2} = 0$

${x}^{2} + x + 1 = 0$

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \times 1 \times 1}}{2 \times 1}$

$x = \frac{- 1 \pm \sqrt{- 3}}{2}$

$\therefore$There are no vertical asymptotes.

Hence, the critical numbers are $x = 0$ and $x = - 2$.

Hopefully this helps!