2

sketch of f(x)

sketch of 1/f(x)

First picture is $y=f(x)$, second picture is $y=\frac{1}{f(x)}$

For this type of question, i have never understood how to determine the concavity of the reciprocal function? From $0<x<a$, there is a point of inflexion, but how does the original qraph gives us this. If anything, the $\cap$-looking curve in the original graph would suggest to me that the reciprocal should also be symmetrical between $a$ and $-a$ but yet they clearly aren't.

Gnolius
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  • Plot $y=f(x)$ on (thin) paper, turn the paper over and rotate so that the positive y axis is now where the positive x axis should be. This is a graph of $x=f^{-1} (y)$ – Paul Jul 13 '23 at 15:42
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    could you elaborate abit more, I understand that the inverse function is the turned over and rotated original function, but i don't quite see the link between the inverse and the reciprocal – Gnolius Jul 13 '23 at 15:48
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    I agree with you that the function looks symmetric on [-a,a], and that its reciprocal should therefore be symmetrical too. I see no reason to assume that the reciprocal is smooth at (-a,0) and (a,0). It could well be non-differentiable there, having different slopes on either side of each of those points. Without knowing in advance that the reciprocal is smooth there, you could hardly expect to judge it from the graph of f. – Jaap Scherphuis Jul 13 '23 at 16:12
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    Oops, only scanned qn on my phone, thought it was about the inverse function sorry – Paul Jul 13 '23 at 18:27

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