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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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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