Question

What are the critical points of `f(x)= -x^2 + 3x` on the interval [0,3]?

Answer 1

First of all. You must get the first derivative of f(x) = `-x^2 + 3x`

The first derivative is `f^'(x)` = -2x+3.

Now you must set `f^'(x)` = 0

You must also find where `f^'(x)` does not exist. But since `f^'(x)` is a polynomial, it is defined everywhere.

`f^'(x)` = 0

-2x+3 = 0

x = 3/2

Since 3/2 is define in the original function, it is a critical number.
X = 3/2 is not rejected because it exist in the closed interval.