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

Question

Scarlett Allison · Accepted Answer

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.

William Abdalla · Answer

undefined To find the extrema of $ h(x) = 7x^5 - 12x^3 + x $, we need to find the critical points where the derivative is zero or undefined.

1. Take the derivative of $ h(x) $ with respect to $ x $.
$ h'(x) = 35x^4 - 36x^2 + 1 $

2. Set the derivative equal to zero and solve for $ x $ to find the critical points.
$ 35x^4 - 36x^2 + 1 = 0 $

3. This equation can be solved using methods like factoring, the quadratic formula, or numerical methods. After solving, you'll find the critical points.

4. Once you have the critical points, you can determine which of them correspond to local extrema by checking the sign of the second derivative or by considering the behavior of the function around those points.

5. Additionally, check the behavior of the function as $ x $ approaches positive and negative infinity to determine if there are any absolute extrema.