Sketch the graph of a continuous function $y=h(x)$ such that

$$

\begin{array}{l}{\text { a. } h(0)=0,-2 \leq h(x) \leq 2 \text { for all } x, h^{\prime}(x) \rightarrow \infty \text { as } x \rightarrow 0^{-}} \\ {\text { and } h^{\prime}(x) \rightarrow \infty \text { as } x \rightarrow 0^{+}} \\ {\text { b. } h(0)=0,-2 \leq h(x) \leq 0 \text { for all } x, h^{\prime}(x) \rightarrow \infty \text { as } x \rightarrow 0^{-}} \\ {\quad \text { and } h^{\prime}(x) \rightarrow-\infty \text { as } x \rightarrow 0^{+} \text { . }}\end{array}

$$

## Discussion

## Video Transcript

All right, this is chapter two section for problem number 18. And for this one, uh, we're just required to sketch a function. Um, and for this function, we want d ah, first derivative of the function to always be positive for the second derivative toe. Always be negative when X is less than two. And to always be positive when X is more than two. Um, so for this first derivative, that's not anything. Doesn't put any huge restrictions on a graph. It just always has to be increasing. So it could look like that line. It could look like that. It could swoop downwards like that. There are a lot of options for this one. Now, for the second derivative, it just means it has to be negative. So going downwards, less than two positive. So curving upwards. Greater than two. Um, except in this case, we can't have these segments that I just crossed out because during those segments, um, if the first derivative is negative and first we want the first rivers. You positive? So, actually, if I redraw those, we're here. We can't have that. We can't have that. So if we put our two more right there. Okay, We have this positive section of the curve right here. We can drag that over and put it to the right of to. And this satisfies our thing so far, because the first derivative is positive. It's increasing. And the second driven to this positive because it's curving up and X is greater than two. Um, and we can take this other part over here and put this on the left of to right, because this is a negative second derivative, because its curving down, which is good because it's less than two, and it is still increasing as well. It never has a negative slope. Um, so this is a function that fits the criteria were looking for, um, So there you have it again. It's always positive and increasing. Um, the second derivative. His negative blow to and positive above two. Here you go.

## Recommended Questions

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