Question

find the maximum, minimum and point of inflexion of f(x)= $\frac{x^3}{3} - x^2$

Answer 1

Answer:

y = x³/3 - x²

dy/dx = 3x²/3 - 2x

y' = x²-2x

put y' = 0

=> x²-2x = 0

=> x(x-2) = 0

x = 0 and x = 2 are the points where the slope is zero. these points are either max or min

let's take the point to the left of x = 0

at x = -1, y' = (-1)² - 2(-1) = 1 + 2 = 3. as the slope to the left of the point at x= 0 is positive, it is increasing hence at x = 0, we get local max of the curve.

put x = 0 in the original function to find the corresponding value of y

0³/3 - 0² = 0

therefore maximum value at x = 0 is 0

now let's take a point to the left of x= 2

at x = 1, y' = (1)² - 2(1) = - 2, so as the slope to tne left of x= 2 is negative, it is a decreasing.

so at x=2 we get local minimum.

therefore the minimum value is obtained by substituting x=2 in the original function of

x³/3 - x²

=> 2³/3 - 2²

=> 8/3 - 4

=> (8 - 12)/3

=> -4/3

therefore min value is - 4/3 or

- 1.33 at x = 2