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Math Help - What is the inverse of this function!?

  1. #1
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    What is the inverse of this function!?

    f(x) = tan(x) +3x for x in (-pi/2,pi/2)

    This is just a little part of the whole question which is on the Inverse Function Theorem, I need to find the inverse to find out if it can be differentiated but can't seem to find the inverse to begin with! Please help!
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    Quote Originally Posted by nlews View Post
    f(x) = tan(x) +3x for x in (-pi/2,pi/2)

    This is just a little part of the whole question which is on the Inverse Function Theorem, I need to find the inverse to find out if it can be differentiated but can't seem to find the inverse to begin with! Please help!
    You aren't going to be able to write the inverse function explicitly.

    Please post the whole question as I'm sure there is something missing...
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    Prove that the function f(x) = tan x +3x for x in (-pi/2, pi/2) has an inverse function that is differentiable on R, and find the value of the derivative of the inverse at 0.

    I have checked that the conditions for the IFT hold and I think they do, so I thought I needed to find the inverse at this point?! Thank you for helping!
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    The function f(x) = \tan{x} + 3x is invertible because it is strictly increasing, (the derivative is f'(x) = \sec^2{x} + 3, which is always positive.

    The function f(x) is differentiable everywhere x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).

    Therefore its inverse will also be differentiable everywhere except where f'(x) = 0.

    If f'(x) = 0

    \sec^2{x} + 3 = 0

    \sec^2{x} = -3.

    This does not have a solution in the real numbers, so f'(x) \neq 0, and so f^{-1}(x) exists and will be differentiable.



    The derivative of the inverse is given by the Inverse Function Theorem:

    \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(x)}

     = \frac{1}{\sec^2{x} + 3}

     = \frac{1}{\frac{1}{\cos^2{x}} + 3}

     = \frac{1}{\frac{1 + 3\cos^2{x}}{\cos^2{x}}}

     = \frac{\cos^2{x}}{1 + 3\cos^2{x}}.
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    Oh i see!! I just needed to put f'(x) into the formula like that, thank you so much for that! I understand it now! Thank you!
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    Quote Originally Posted by Prove It View Post
    The function f(x) = \tan{x} + 3x is invertible because it is strictly increasing
    That suffices? What about f:[0,1]\mapsto\mathbb{R} given by f(x)=x?
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    Quote Originally Posted by Drexel28 View Post
    That suffices? What about f:[0,1]\mapsto\mathbb{R} given by f(x)=x?
    Why wouldn't this function be invertible? It's a self-inverse function after all...
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Prove It View Post
    Why wouldn't this function be invertible? It's a self-inverse function after all...
    A function has an inverse if and only if it's bijective. f(x)\ne\pi,\text{ }x\in[0,1]
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