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

Question

3379 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-05T18:36:57

Extrema are at x=#+-1# and x=#+-sqrt(1/35)#

h(x)= #7x^5 -12x^3 +x#

h'(x)= #35x^4 -36x^2 +1#

Factorising h'(x) and equating it to zero, it would be#(35x^2 -1)(x^2-1)=0#

The critical points are therefore #+-1, +-sqrt(1/35)#

h''(x)= #140x^3-72x#

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(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.