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I know that h and h'' have to be one of those curves due to the concavity of f(x) at f''(x). However I am having difficulty understanding how am I supposed to know what is h and what is h''.

I think h and h'' should be switched in picure on the left. Or is the solution correct?

enter image description here

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  • $\begingroup$ To make this a good question, please add why you think they should be switched. $\endgroup$ Nov 9, 2021 at 19:29
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    $\begingroup$ I hope whoever marks the questions will accept both the solution as shown and the one where $h$ is swapped with $h''$ and then the ones where $k$ is potentially swapped with $k''$ as well. Namely, without the scale on the $x$ and $y$ axis it is impossible to say which of the two functions is the second derivative of the other, if they are both sine-like functions. Notice e.g. $\frac{d^2}{dx^2}(\sin x)=-\sin x$ and $\frac{d^2}{dx^2}(-\sin x)=\sin x$ $\endgroup$ Nov 9, 2021 at 19:32

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As already noted in the comments, without scale marks on the coordinate axes, it is actually impossible to tell.

For instance, here is a plot of $$\color{blue}{y = k(x) = \sin \frac{x}{2}}, \\ \color{red}{y = k''(x) = -\frac{1}{4} \sin \frac{x}{2}},$$ on the interval $x \in [-4\pi, 4\pi]$:

enter image description here

Notice how the second derivative has amplitude less than the original function.

Now consider instead

$$\color{green}{y = k(x) = -\sin 2x}, \\ \color{purple}{y = k''(x) = 4\sin 2x},$$

on the interval $x \in [-\pi, \pi]$:

enter image description here

Other than the scale markings, the plots are (nearly) identical in appearance but switched. Hence you cannot tell which is which unless you are also given the scale markings.

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