# Calc.

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Let h(x)=16-4x-(x^3)
Find g(96).

My work:
96= 16-4x-(x^3)
(-x^3)-4x-80=0
Using the rational roots theorm, I found that x=-4
so h(-4)=96, therefore g(96)=-4.

Find g'(96).

My work:
g'(96)= 1/(dh/dx at x=-4)
1/((-3(-4)^2)-4)= -1/52

Does this look ok? Also, my math teacher wanted us to find two ways to do this problem and get the same answer. I found one way, but I can't think of another. Any suggestions?

• Calc. -

Even though you did not state it, I can see from your work that g(x) must be the inverse of h(x)

the alternate way suggested by your teacher might have gone like this:

h(x) = 16-4x-(x^3) is equivalent to
y = 16-4x-(x^3)

then g(x) would be equivalent to
x = 16 - 4y - y^3
I then differentiated this implicitly to get
dy/dx = -1/(4+3y^2)
but remember (-4,96) and (96,-4) are inverse images, so for g'(96) we would use y=-4 in dy/dx to get -1/(4 + 3(16))
= -1/52 as before

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