Question

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

Answer 1

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

Explanation:

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.