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Problem 66

Sketch the graph of a differentiable function $y=…

Problem 65

Sketch the graph of a differentiable function $y=f(x)$ through the
point $(1,1)$ if $f^{\prime}(1)=0$ and
\begin{array}{l}{\text { a. } f^{\prime}(x)>0 \text { for } x<1 \text { and } f^{\prime}(x)<0 \text { for } x>1} \\ {\text { b. } f^{\prime}(x)<0 \text { for } x<1 \text { and } f^{\prime}(x)>0 \text { for } x>1} \\ {\text { c. } f^{\prime}(x)>0 \text { for } x \neq 1} \\ {\text { d. } f^{\prime}(x)<0 \text { for } x \neq 1}\end{array}


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Video Transcript

we have four graphs to draw here in situation A. We have a graph that has a point of 11 And in all these grafts, we actually have a point of 11 on the derivative at that point is zero. So it's a horizontal slope and zero in part A. When X is below one, the slope should be positives. That's gonna be an increasing slope or encouraging function before one. And then afterwards a decreasing functions. So negative slope. So one way to do that would be to have it look like a parabola opening downwards. In the second graph again, we have the 0.11 slope of zero. At that point prior to one. The slope is negative. So it's a decreasing function afterward and positive, so that could look like oppa rabble that opens up four part C. We again have a point of 11 slope of zero there, and the derivative, as long as X isn't one is positive. So we need it increasing before it gets to one. An increasing after, but a slope of zero at one. So what's gonna happen here? Said it levels off and then flips over and finally, in part D, we again have a point of 11 The slope to its left is negative. So it's a decreasing function, the slope of the rightist negative. So this time it will come down that little off and then go down.

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