Question

Question #115fc

Answer 1

Maximum at `(-8,75)`

Explanation:

We have:

`g(x)=-x^2-16x+1`

Whose graph is as follows:
graph{-x^2-16x+1 [-25, 10, -20, 80]}

General Observations

`g(x)` is a quadratic so we have a parabola with a single turning point. The `x^2` coefficient is negative so the parabola is inverted (`nn` shaped) and so we have a single maximum .

We can demonstrate this is the case using two methods:

Method 1 - Completing the Square

`g(x) = -(x^2+16x-11)`
`" " = -((x+16/2)^2-(16/2)^2-11)`
`" " = -((x+8)^2-8^2-11)`
`" " = -((x+8)^2-64-11)`
`" " = -((x+8)^2-75)`
`" " = -(x+8)^2+75`

Clearly `(x+8)^2 ge 0 AA x in RR`, and `x+8=0=>x=-8` and so `g(x)` has a maximum of `75` when `x=-8`

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 `x=-8`

The nature of the critical point is determined by the sign of the second derivative:

`x=-8 => g''(-8) = -2 < 0`

And as `g''(x) < 0` the critical point is a maximum .