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

Mar 28, 2015

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

The first derivative is ${f}^{'} \left(x\right)$ = -2x+3.

Now you must set ${f}^{'} \left(x\right)$ = 0

You must also find where ${f}^{'} \left(x\right)$ does not exist. But since ${f}^{'} \left(x\right)$ is a polynomial, it is defined everywhere.

${f}^{'} \left(x\right)$ = 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.