# What are the extrema of h(x) = 7x^5 - 12x^3 + x?

Jul 5, 2016

Extrema are at x=$\pm 1$ and x=$\pm \sqrt{\frac{1}{35}}$

#### Explanation:

h(x)= $7 {x}^{5} - 12 {x}^{3} + x$

h'(x)= $35 {x}^{4} - 36 {x}^{2} + 1$

Factorising h'(x) and equating it to zero, it would be$\left(35 {x}^{2} - 1\right) \left({x}^{2} - 1\right) = 0$

The critical points are therefore $\pm 1 , \pm \sqrt{\frac{1}{35}}$

h''(x)= $140 {x}^{3} - 72 x$

For x=-1, h''(x)= -68, hence there would be a maxima at x=-1

for x=1, h''(x)= 68, hence there would be a minima at x=1

for x=$\sqrt{\frac{1}{35}}$, h''(x) = 0.6761- 12.1702= - 11.4941, hence there would be a maxima at this point

for x= #-sqrt (1/35), h''(x) =-0.6761+12.1702=11.4941, hence there would be a minima at this point.