For what value of x will the function f(x) = -3(x - 10)(x - 4) have a maxim

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We have the function f(x) = -3(x - 10)(x - 4). to find the value where this has a maxima, we need to find the first derivative, equate it to zero and solve for x.

f(x) = -3(x - 10)(x - 4)

=> f(x) = -3( x^2 - 14x + 40)

=> -3x^2 + 42x - 120

f'(x) = -6x + 42

f'(x) = 0

=> -6x + 42 = 0

=> x = 42/ 6 = 7

Also f''(x)  = -6, which...

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neela | Student

Q:For what value of x will the function f(x) = -3(x - 10)(x - 4) have a maximum.

A:

f(x) = -3(x-10)(x-4)

f(x) = -3(x^2-4x-10x+40)

f(x) = -3(x^2-14x+40.

f(x) = -3{(x-7)^2 - 7^2+40}

f(x) = - 3{(x-7)^2 -9}

f(x) = 27  - 3(x-7)^2.

So (x-7)^2, being a perfect square is always > 0  for all x and  (x-7) ^2 = 0 for x= 7.

Therefore f(x) = 27-3(x-7)  = 27 is the maximum for x= 7.

So f(x) = -3(x-10)(x-4) is maximum when x = 7, and maximum f(x) = f(7) = 27.

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