Question

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

Accepted Answer

Given the function f(x) = -3(x-10)(x-4)
We need to find the maximum value.
First we will open the brackets and expand the function.
==> f(x) = -3(x^2 -14x + 40)
               = -3x^2 + 42x - 120
==> f(x) = -3x^2 + 42x - 120
Now we need to find the first derivative.
==> f'(x) = -6x + 42
Now we need to find the critical values which is the derivatives zeros.
==> -6x + 42 = 0
==> -6x = -42
==> x = 7
Then the function has a maximum values when x = 7
f(7) = -3(7-10)(7-4) = -3*-3*3 = 27
The maximum values of f(x) is the point ( 7, 27)

Answer

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.

Answer

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 is negative at x = 7
At x = 7, f(x) = -3( -3)( 3) = 27
Therefore the maxima is at x = 7 and the maximum value is 27.

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

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