Find the maximum or minimum value of f(x) = -3x^2 + 9x
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calendarEducator since 2010
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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
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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)
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.
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