What is the minimum value of 4x^2-8x-18 ?
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To find the minimum value of y = 4x^2 - 8x -18, we need to find the first derivative and equate it to 0. We solve for x and determine the value of the function for the value of x that we find.
y = 4x^2 - 8x - 18
y' = 8x - 8
8x - 8 = 0
=> x = 8/8
=> x = 1
Also y'' = 8 which is always positive , therefore at x = 1, we have the minimum value.
y at x = 1,
=> 4*1^2 - 8*1 - 18
=> 4 - 8 - 18
=> -22
The minimum value of 4x^2 - 8x -18 is -22.
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To find out the extreme value of a function, we'll have to calculate the first derivative of the function.
Let's find the first derivative of the function f(x):
f'(x)=( 4x^2-8x-18)'=(4x^2)'-(8x)'-(18)'
f'(x)=8x-8
Now we have to calculate the equation of the first derivative:
8x-8=0
We'll divide by 8:
x-1 = 0
x=1
That means that the function has an extreme point, for the critical value x=1.
f(1) = 4*1^2 - 8*1 - 18
f(1) = 4 - 8 - 18
We'll combine like terms:
f(1) = -22
The extreme point of the function is a minimum point whose coordinates are: (1 ; -22).
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