Calculus Problem!!! Find the value of df^-1/dx at x = f(a)?

Question

Find the value of df^-1/dx at x = f(a)?f(x) = x^3 - 9x^2 - 4, x ≥ 6, a = 3 I Think you find the derivative and get: 3x^2 - 18xf(x)^-1 = 1/ (3x^2 - 18x)----------f(3)= (3)^3 - 9(3)^2 - 4 = -58 -----------so.... 1/(3(-58)^2 - 18(-58)) = 1/11136That seems way off but I dont really know. Please help me where I went wrong or how to do it!

Mr. Toni Zulauf Sr. · Accepted Answer

f'(x)=3x^2-18x, then(df/dx)^-1=1/(3x^2-18x), so(df/dx)^-1=1/(27-54)=-1/27But another point to consider is that f^-1 may denote the inverse of f:f^-1 != 1/ff^-1 denotes the inverse of fAccording to the inverse function theorem:(d/dx)f^-1(a)=1/f'(b), where a=f(b)We want (d/dx)f^-1(3), so 3=f(b), so3=x^3-9x^2-4 -> 0=x^3-9x^2-7You can use the cubic formula to solve the above eqn. I won't do it for you because it's a pain. Then you plug your solution into (d/dx)f^-1(a)=1/f'(b) I don't feel a strong urge to do the analysis, but it appears to me that f^-1 is probably not a function.Regardless, you'll get 2 complex roots and 1 real root from the cubic formula. The real root is b=9.0848 which gives 1/f'(b)=1/84.074 which may or may not be (d/dx)f^-1(3) depending on whether f^-1 is a function

Sigrid O'Kon

Calculus Problem!!! Find the value of df^-1/dx at x = f(a)?

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