Given the curve f(x) = -3x^2 + 9x.

We need to find the extreme value of the function.

First we notice that the coefficient of x^2 is negative, then the curve will have a maximum point.

Now we will find the first derivative.

=> f'(x) = -6x + 9

Now we will determine the critical value which is the derivatives zero.

==>< -6x + 9 = 0

==> x = -9/-6 = 9/6 = 3/2

==> x = 3/2

Now we will calculate f(3/2)

==> f(3/2) = -3(3/2)^2 + 9(3/2) = -27/4 + 27/2 = 27/4

**Then the function f(x) has a maximum value at the point (3/2, 27/4)**

We find the extreme points of f(x) by finding its derivative and equating that to zero. This is then solved to determine the zeros.

f(x) = -3x^2 + 9x

f'(x) = -6x + 9

-6x + 9 = 0

=> -6x = -9

=> x = 9/6

=> x = 3/2

At x = 0, f(x) = -3*(3/2)^2 + 9*(3/2)

=> -3( 9/4) + 27 / 2

=> -27 / 4 + 27/2

=> 27/4

f''(x) = -6 which is negative for x = 3/2.

**Therefore the maximum value of f(x) = -3x^2 + 9x = 27/4**

To find the maximum value of f(x) = -3x^2+9x.

Solution:

f(x) = -3x^2+9x = -3{x^2-3x}....(1)

x^2-3x = x^2-3x+(3/2)^2 - (3/2)^2 = (x-3/2)^2 - (3/2)^2.

So f(x) = -3(x^2-9x) = -3{(x-3/2)^2 -9/4)}.

=> f(x) = 27/4 - 3(x-3/2)^2. Here (x-3/2)^2 is perfect square. So (x-3/2)^2 >= 0. So 27/4 - 3(x-3/2)^2 is always <= 27/4.

So f(x) = -3x^2+9x = 27/4- 3(x-3/2)^2 < = 27/4. So f(x) = 27/4 is the maximum um value when x= 3/2.

**f(x) = -3x^2+9x has the maximum value 27/4, when x = 3/2**.