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

My work:
96= 16-4x-(x^3)
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)
    Both of your answers are correct.

    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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