# Is the solution to picture on the left correct? [closed]

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? • To make this a good question, please add why you think they should be switched. Nov 9, 2021 at 19:29
• 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 $\frac{d^2}{dx^2}(-\sin x)=\sin x$ Nov 9, 2021 at 19:32

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 $$x \in [-4\pi, 4\pi]$$: Notice how the second derivative has amplitude less than the original function.

$$\color{green}{y = k(x) = -\sin 2x}, \\ \color{purple}{y = k''(x) = 4\sin 2x},$$
on the interval $$x \in [-\pi, \pi]$$: 