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

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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) **

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.**

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.**