Question

Find extreme values of function f(x) = x³ - (3/2)x² on interval [- 1,2]

Answer 1

-2 decimal 3/% because if you divide the principal in half its 1% of 3

Answer 2

we have the function

`f(x)=x^3-(3)/(2)x^2`

Find out the first derivative of the given function

`f^(\prime)(x)=3x^2-3x`

Equate to zero the first derivative

`\begin{gathered} 3x^2-3x=0 \\ 3x(x-1)=0 \end{gathered}`

The values of x are

x=0 and x=1

we have the intervals

(-infinite,0) (0,1) (1,infinite)

Interval (-infinite,0) -----> f'(x) is positive

interval (0,1) ---------> f'(x) is negative

interval (1,infinite) -----> f'(x) is positive

that means

x=0 is a local maximum

x=1 is a local minimum

Find out the y-coordinates of the extreme values

For x=0 -----> substitute in the function f(x) ---------> f(x)=0

For x=1 ------> substitute in the function f(x) ------> f(x)=-0.5

therefore

The extreme values are

local maximum at (0,0)

local minimum at (1,-0.5)