# What are the local extrema of f(x)= (x^5-x^2-4)/(x^3-3x+4)?

Feb 10, 2016

#### Answer:

Local maximum $\approx - 0.794$ (at $x \approx - 0.563$) and local minima are $\approx 18.185$ (at $x \approx - 3.107$) and $\approx - 2.081$ (at $x \approx 0.887$)

#### Explanation:

$f ' \left(x\right) = \frac{2 {x}^{7} - 12 {x}^{5} + 21 {x}^{4} + 15 {x}^{2} - 8 x - 12}{{x}^{3} - 3 x + 4} ^ 2$

Critical numbers are solutions to

$2 {x}^{7} - 12 {x}^{5} + 21 {x}^{4} + 15 {x}^{2} - 8 x - 12 = 0$.

I do not have exact solutions, but using numerical methods will find real solutions are approximately:

$- 3.107$, $- 0.563$ and $0.887$

$f ' ' \left(x\right) = \frac{2 {x}^{9} - 18 {x}^{7} + 14 {x}^{6} + 108 {x}^{5} - 426 {x}^{4} + 376 {x}^{3} + 72 {x}^{2} + 96 x - 104}{{x}^{3} - 3 x + 4} ^ 3$

Apply the second derivative test:

$f ' ' \left(- 3.107\right) > 0$, so $f \left(- 3.107\right) \approx 18.185$ is a local minimum

$f ' ' \left(- 0.563\right) < 0$, so $f \left(- 0.563\right) \approx - 0.794$ is a local maximum

$f ' ' \left(0.887\right) > 0$, so $f \left(0.887\right) \approx - 2.081$ is a local minimum