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

Expert Answers info

justaguide eNotes educator | Certified Educator

calendarEducator since 2010

write12,544 answers

starTop subjects are Math, Science, and Business

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

check Approved by eNotes Editorial
hala718 eNotes educator | Certified Educator

calendarEducator since 2008

write3,662 answers

starTop subjects are Math, Science, and Social Sciences

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)

check Approved by eNotes Editorial

neela | Student

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.

check Approved by eNotes Editorial