Question #115fc
1 Answer
Maximum at
Explanation:
We have:
g(x)=-x^2-16x+1 g(x)=−x2−16x+1
Whose graph is as follows:
graph{-x^2-16x+1 [-25, 10, -20, 80]}
General Observations
We can demonstrate this is the case using two methods:
Method 1 - Completing the Square
g(x) = -(x^2+16x-11) g(x)=−(x2+16x−11)
" " = -((x+16/2)^2-(16/2)^2-11) =−((x+162)2−(162)2−11)
" " = -((x+8)^2-8^2-11) =−((x+8)2−82−11)
" " = -((x+8)^2-64-11) =−((x+8)2−64−11)
" " = -((x+8)^2-75) =−((x+8)2−75)
" " = -(x+8)^2+75 =−(x+8)2+75
Clearly
Method 2 - Calculus
Differentiating (twice) we get:
\ g'(x)=-2x-16
g''(x)=-2
At a critical point (min or max) the first derivative vanishes, thus:
g'(x)=-2x-16 = 0 => -2x-16 = 0
:. 2x=-16 => x=-8
So there is one critical point when
The nature of the critical point is determined by the sign of the second derivative:
x=-8 => g''(-8) = -2 < 0
And as
