if a question asks for just its maximum value do you consider the end points?OR only its Local max&min values?

Question

for a quadratic equation: x^3 - 4x^2 it has its maximum value on interval [-2,5] at x=0
- is that true? or its maximum value would be its end point at x=5?

Answer 1

Definitely consider the endpoints! In this case 25 > 0, so the maximum on that interval is at (5, 25).

Answer 2

Yes, you should check the end-points to get the minimum and maximum:

If you derive your function to find them you'll find that the derivative is zero at two points

3x^2 - 8x = 0

3x(x - 8/3) = 0

8/3 = 2+2/3


x=0,8/3 both are within the interval

Let's derive again:

6x - 8 = 0

When x = 0 the 2nd derivative is -8, so 0 is a local maximum, at -2 we can find the minimum point.

When x = 8/3 6x - 8 = 6*8/3 - 8 = 12 - 8 = 4 > 0

8/3 is a local minimum point.

(-2)^3 - 4(-2)^2 = -8 - 4*4 = -8-16 = -24

0^3-4*0^2 = 0

Now (8/3)^3 - 4(8/3)^2 = 512/27 - 4*64/9 = 512/27 - 256/9 =
= 512/27 - 768/27 = -256/27 = -(9 + 13/27)

5^3-4*5^2 = 125 - 4*25 = 125 - 100 = 25

Taking maximum and minimum from the 4 values found:

-24 is the minimum over the interval, and 25 is the maximum.
Both values are achieved at the endpoints.

Answer 3

The question is to find the maximum value of the function between -2 and 5, including the endpoints. If the interval specified used parthentheses (-2, 5) then the endpoints would not be included. The brackets indicate that the endpoint values are included in the interval.

Answer 4

You must consider end points, unless you are specifically asked for a local maximum turning point.

Likewise, in this case although there is a local minimum of roughly -9.5 at x = 8/3, the minimum value -24 is at the left end.