For Function F(x)= x^3-75x?

Question

find:

Local Maxima ; which occurs

Local Minima ;which occurs

Answer 1

Use the first and second derivative tests.
First, look for critical points of the function finding when the first derivative is 0

F(x) = x^3-75x
F'(x) = 3x^2 -75
F'(x) = 3(x^2 - 25)
F'(x) = 3(x+5)(x-5)
0 = 3(x+5)(x-5)

So x= 5 and x= -5 are critical points. Next Find out whether the function is concave down or up at those points to figure out if it's a max or a min, respectively. By checking out the sign of the second derivative at those points.

F''(x) = 6x
F''(5) = 30 concave up U so x =5 is a local min
F"(-5) = -30 concave down ∩ so x=-5 is a local max

local min at x = 5 and local max at x=-5
(5,-250) and (-5,250)

Answer 2

F´(x) = 3x^2 -75
3x^2-75 = 0 x=+-5
the sign of F´(x) is ++++++(-5)--------(5)+++++++++
so at x=-5 we have a local maximum and at x=5 a local minimum
F(-5)=250
F(5) = -250

Answer 3

3x^2-75=d(f(x))/dx
3(x-5)(x+5)
so at +5 and -5 are where the slope is zero
F(-5) = -125 + 375 = 250
F(5) = 125 - 375 = -250

Answer 4

what he said