Question #115fc

1 Answer
Jul 14, 2017

Answer:

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 .