Question

Use the function f(x) = x^4 - 36 x^3

At what value(s) of x does f have a local maximum or a local minimum?

Answer 1

x=0 at the slope of the curve but i couldnt find the other point

Answer 2

dy/dx = 4x^3 - 108x^2
when dy/dx = 0, (That means maximum and minimum)
4x^3-108x^2 =0
x^2 (4x-108)=0
x=0 x=27

Answer 3

dy/dx = 0 because when maximum or minimum point, the gradient is 0. and dy/dx is gradient.

Answer 4

f'(x) = 4x^3 - 108x^2 = 0
-> 4x^2 ( x - 27) = 0
=> x = 0, x = 27

Answer 5

x = 0, y = 0 =>(0,0) Max
x = 27, y = 27^3 ( 27 - 36) = -177147 ->( 27, -177147) min

Answer 6

x^3 (x-36) = 0

x^3 = 0 or x = 36

Triple root at 0 indicates that graph is tangent to x-axis --> neither max nor min

Graph slices through x-axis at 36

Dominant term of x^4 indicates entry into Quad 2 and exit in Quad 1

Absolute minimum at x = 27 --> (27, - 177,147) (needs checking)