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Let $f(x) = 2x + \cos{x}$.

  • Calculate $f^{-1}(1)$;
  • Supposing that $f^{-1}$ is differentiable, show that $(f^{-1})^{\prime}(1) = \dfrac{1}{2}$.

Thanks for the help :]

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    Draw the graph on a graphic calculator to find $f^{-1}(1)$. For the derivative, use the formula for derivative of $f^{-1}$. – 2017-02-17
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    @MaadhavGupta It's not a duplicate. He's not asking for an explicit formula for the inverse. – 2017-02-17
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    see here i hope this will help you https://en.wikipedia.org/wiki/Inverse_functions_and_differentiation – 2017-02-17

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We first need to confirm whether $f^{-1}(x)$ exists. To do so, we need to show it's a bijection. Let $x_1,x_2$ be such that $$f(x_1)=f(x_2) , x_1 \ne x_2$$ Hence, $$2x_1 + \cos x_1 = 2x_2 + \cos x_2$$ $$\implies \frac{\cos x_1 - \cos x_2}{x_1 - x_2} = -2$$ From LMVT, this is not possible , as the LHS must be $-\sin x$ for some point $x$, which must have absolute value less than $1$. Hence, $f(x_1) = f(x_2) \implies x_1 = x_2$

Hence, function is invertible. Now that $f^{-1}(x)$ exists, we calculate it for x = 1. This is equivalent to solving the equation $$2x + \cos x = 1$$ With a bit of rearrangement, we have $$x = \sin ^{2} \frac{x}{2}$$ One solution is $x = 0$. Since it is a bijection, this must be the only solution. Hence, $f^{-1}(1) = 0$

For the second part, we use the fact that $$f(f^{-1}(x)) = x$$ Differentiating both sides, and given that all required functions are differentiable, we have $$f'(f^{-1}(x)) * (f^{-1})'(x) = 1$$ For $x = 1$, this returns $$f'(0) * (f^{-1})'(1) = 1$$ $$\implies (f^{-1})'(1) = \frac{1}{2}$$