a tough calculus problem?

Question

any help would REALLY be appreciated

f(x)= x^(4/3) + 4x^(1/3) on the interval -8<=x<=8

a) find the coordinates of all points at which the tangent line to the curve is horizontal.

b) find the coordinates of all points at which the tangent line to the curve is vertical.

c)find the coordinates at whcih the absolute maximum and absolute minimum occur.

d) for what values of x is the function concave down?

thank you so much for any help.

Answer 1

a. f'(x) = (4/3)x^(1/3)+(4/3)*x^(-2/3) = 0 for x = -1. This is the only point where the tangent is horizontal.
b. from a. f'(x) = infinite for x = 0 making the tangent vertical.
c. from a. follows the extreme max/min at x = -1. Since f''(x) = (1/3)x^(-2/3) - (2/3)x^(-5/3) = 1 > 0 for x = -1 the extreme at x = -1 is minimum.
d. Since within the given interval the function is final and has only one minimum that is not at the end of the interval the function is concave downward (I would say upward) meaning from f(-8) = 8 monotonically decreases to the minimum f(-1) = -3 and monotonically increases again to f(8) = 24 which is then absolute maximum..

n.b. you can go to excell and plot the function.

Answer 2

you should already know the derivative of f(x) from a previous post

a) let f'(x)=0 and solve for x
b) where is f'(x) undefined but f(x) IS defined
c) find f(x) for all the values of x that you found in part a, as well as the endpts of the interval -8 and 8
d) take the second derivative of f(x) and determine when it is negative.