# Question #115fc

##### 1 Answer

#### Answer:

Maximum at

#### 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**

**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

**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 **maximum** .