What are the extrema of h(x)=7x512x3+x?

1 Answer
Jul 5, 2016

Extrema are at x=±1 and x=±135

Explanation:

h(x)= 7x512x3+x

h'(x)= 35x436x2+1

Factorising h'(x) and equating it to zero, it would be(35x21)(x21)=0

The critical points are therefore ±1,±135

h''(x)= 140x372x

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=135, 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.