# Find the maximum or minimum value of f(x) = -3x^2 + 9x

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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)

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.