logan.sowder38
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- Oct 21, 2015
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\(\displaystyle 1.\, \mbox{ If }\, f(x)\, =\, \dfrac{x^3}{x^2\, +\, 1},\, \mbox{ find }\, \left(f^{-1}\right)' (2)\)
Here's my work so far.
\(\displaystyle f(x)\, =\, \dfrac{x^3}{x^2\, +\, 1}\). . . . .\(\displaystyle \mbox{Find }\, \left(f^{-1}\right)' (2)\)
\(\displaystyle \mbox{Let }\, g(x)\, =\, f^{-1}(x)\, \longrightarrow \, \mbox{ find }\, g'(2)\)
\(\displaystyle g'(x)\, =\, \dfrac{1}{f'(g(x))}\)
I understand to use the theorem (f-1)' (a) = 1/(f'(f-1(a)), but I can't figure out how to solve for x so I can plug it into f'(x)
Here's my work so far.
\(\displaystyle f(x)\, =\, \dfrac{x^3}{x^2\, +\, 1}\). . . . .\(\displaystyle \mbox{Find }\, \left(f^{-1}\right)' (2)\)
\(\displaystyle \mbox{Let }\, g(x)\, =\, f^{-1}(x)\, \longrightarrow \, \mbox{ find }\, g'(2)\)
\(\displaystyle g'(x)\, =\, \dfrac{1}{f'(g(x))}\)
\(\displaystyle f(x)\, =\, 2\) \(\displaystyle 2\, =\, \dfrac{x^3}{x^2\, +\, 1}\) \(\displaystyle 2x^2\, +\, 2\, =\, x^3\) \(\displaystyle x^3\, -\, 2x^2\, -\, 2\, =\, 0\, ???\) | \(\displaystyle f'(x)\, =\, \dfrac{(x^2\, +\, 1)(3x^2)\, -\, (x^3)(2x)}{(x^2\, +\, 1)^2}\) \(\displaystyle f'(x)\, =\, \dfrac{3x^4\, +\, 3x^2\, -\, 6x^3}{(x^2\, +\, 1)^2}\) \(\displaystyle f'(x)\, =\, \dfrac{3x^4\, -\, 6x^3\, +\, 3x^2}{x^4\, +\, 2x^2\, +\, 1}\) |
I understand to use the theorem (f-1)' (a) = 1/(f'(f-1(a)), but I can't figure out how to solve for x so I can plug it into f'(x)
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