Question

What is the max value of the derivative of f9x) = 3x^2-x^3 ?

Answer 1

Given:

y = x^3 - 3x^2 + 2

Therefore:

dy/dx = 3x^2 - 3*2x = 3x(x - 2)

Equating above expression for dy/dx to 0, we get two values of x.

x 1 = 0 and x2 = 2

To get corresponding value of function, we substitute these values of y in equation for y. Thus:

y1 = 0^3 - 3*0^2 + 2

= 2

y2 = 3^3 - 3*2^2 + 2

= 27 - 12 + 2

= 17

Answer:

Maximum = 17

Minimum = 2

Answer 2

Hmm.. the answer key says the minimum is 3, but I do not know how