Given the function: h(x) = 5x-3x^2

We need to find the extreme values for h(x)

First, we need to find the first derivative h'(x).

==> h'(x) = 5 - 6x

Now we will determine the derivatives zero.

==> 5 - 6x = 0

==> x = 5/6

Then, the function h(x) has an extreme values when x = 5/6

==> h(5/6) = 5(5/6) - 3(5/6)^2

= 25/6 - 3(25/36

= 25/6 - 75/36 = 75/36 = 25/12

==> h(5/6) = 25/12

We also notice that the sign of x^2 is negative.

**Then, the function has maximum values at the point ( 5/6, 25/12)**

**Or at h(5/6) = 25/12**

To find the extreme value of 5x-3x^2.

Let f(x) = 5x-3x^2

f(x) = {(5/3)x - x^2)3.

f(x) = -3{x^2-(5/3)x}

f(x) = -3 (x^2-(5/3)x+ (5/6)^2} +3*(5/6)^2.. We subtracted and added 3(5/6)^2, without changing the net value of f(x).

f(x) = 25/12 - 3(x-5/6)^2. So we expressed f(x) as a positive quantity - 3 times perfect square which is maximum when x-5/6 = 0.

So f(x) is maximum 25/12 when x= 5/6.

So f(x) attains maximum of 25/12 when x = 5/6.