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Make a careful sketch of the graph of $f$ and below it sketch the graph of $f'$ in the same manner as in Exercises 4-11. Can you guess a formula for $f'(x)$ from its graph?$f(x) = e^x$

The slope at 0 appcars to be 1 and the slope at 1 appears 0. Since to be 2.7 . As $x$ decreases, the slope gets closer to the graphs are so similar, we might gacss that $f^{\prime}(x)=e^{x}$.

Limits

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Anna Marie V.

Campbell University

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

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Video Transcript

this problem for seventeen of the Stuart Calculus eighth edition, Section two point eight make a careful sketch of the graph of F and blew it sketch the graph of every time in the same manner as an exercise is for true love. Can you guess a formula for halftime from its crap? So first, let's plot after of X equals e to the X and the top craft were were sketching Ah, good. The graph of F and looks roughly like this One of the key points is this point zero one, not a word that is the white intercept of the function. Because at X equals zero, it is there is one. Ah, and we see that the limit is expertise needed Infinity a zero and that limit is exported infinities infinity. So this increases. Ah, as a X goes towards infinity. Now I may be a bit difficult, but we're going to attempt to drop the graph of time by using the slopes of the tangent lines that our attention to this function f the easy are one of the easier ones to look at is as X approaches negative infinity. We see that dysfunction that pulls out. Search them. The slopes of the tension line decrease in Stevenage and in value until it becomes almost flat. A sex person. Negative infinity. So we're going to get this Ah, decrease of the slope towards zero. Because the slopes of the tension lines approach zero. As we approach any other affinity arm, we can estimate the slope of the tangent line at X equals zero here and where we end up. If we were to fear that's not exactly the slope of the tension line at X equals zero ends up being exactly one. So there's an interesting result. And what is more interesting is that if we continue to find the slopes of each of the tendon lines as X increases, we will begin. We will continue to plot what is exactly the same graph as F arms. So in reality, what we have just plotted is the functioning of the ex. And this is consistent with our definition for this function, any of the ex because we know that the definition of either the ex is indeed itself even the X and so they're dirt of craft is exactly the same esteem original graph, and it at every point is the slope of the tangent line to dysfunction at every point

Topics

Limits

Derivatives

Anna Marie V.

Campbell University

Kayleah T.

Harvey Mudd College

Caleb E.

Baylor University

Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp