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Find the maximum value of

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f(x) = -3x^2 + 12x-8

asked Mar 16, 2013 in CALCULUS by chrisgirl Apprentice

1 Answer

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The function f(x) = -3x2 + 12x  - 8

The maximum value of  f(x) then the derivative of f(x) = 0

f(x) = -3x2 + 12x  - 8

Diferentiate each side with respective x

Recall :  From derivative formula d / dx(axn) = axn-1

f1(x) = 2(-3x2-1) + 12x1-1 -0

f1(x) = -6x + 12(1)

f1(x) = -6x + 12

Evaluate the maximum value of f(x) then the f1(x) = 0

Substitute f1(x) = 0

There fore 0 = -6x + 12

Add 6x to each side

6x = 6x -6x +12

6x = 12

Divide each side by 6

6x / 6 = 12 / 6

Simplify

x = 2

Substitute x = 2  to f(x) = -3x2 + 12x - 8

f(2) = -3(2)2+ 12(2) - 8

f(2) = -12 +24 - 8

f(2) = -20 + 24

f(2) = 4

There fore the maximum value of  f(x) : f(2) = 4.

answered Mar 16, 2013 by diane Scholar

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