Help with finding the maximum value??

Question

If g(x)=x+3 h(x)=9-x^2, what is the maximum of h(g(x))?

Answer 1

h(g(x)) = 9 - (x+3)^2

 

=9 - (x^2+6x+9)

 

= -x^2 -6x

 

first derivitive = -2x - 6

 

for max or min this is equal to 0

 

hence x = -3

 

h(g(-3)) = (-3)^2 -6(-3) = 9 +18 = 27

 

to prove its a max and not a min second derivative must be negative

 

second derivative = d/dx(-2x-6) = -2 (proven) 

Answer 2

h(x) = 9 - x²   g(x) = x + 3

 

h(g(x)) = h(x+3) = 9 - (x+3)²

 

                        = 9 - (x²+6x+9) = -x² - 6x

 

If y = Ax²+Bx+C and A < 0, the graph is a parabola opening downward with maximum when x = -B/(2A).

 

So, for y = -x²-6x, the maximum value occurs when x=-(-6)/(2(-1))=-3.

 

Maximum value = -(-3)²-6(-3) = 9.