f(x) = 4x5 − 240x3

Question

Find the critical values. (Enter your answers as a comma-separated list.)
x = 1

Find the relative maxima.
(x, y) = (

Find the relative minima.
(x, y) = ( 3

Find the points of inflection. (Round your answers to one decimal place.)
(x, y) = ( 4 ) (smallest x-value)
(x, y) = ( 5 )
(x, y) = ( 6 ) (largest x-value)

Answer 1

First, your expression "4x5 - 240x3" makes no sense unless you're actually using  exponents.  If so, then the proper way to write it out in here and on your TI-89 is:
f(x) = 4*x^5 - 240*x^3

If that's the case then taking the first derivative of f(x) gives:

f'(x) = 20*x^4 - 720*x^2
If you factor that, you will find the local maxima and minima of the function at:
f'(x) = (20*x^2) * (x-6)*(x+6)
The zeroes of this function give us the local maxima and minima as (-6,20736), (0,0) (6,-20736)

To find the inflection points of f(x), take it's 2nd derivative and find the zeros of that function.
f''(x) = 80x^3 + 1440x
factor that into:
f''(x) = (80x)(x^2 - 18) = (80x)(x+sqrt(18))(x-sqrt(18))
So the inflection points occur at x = {-4.2, 0, 4.2}