sketch-the-graph-of-fxxx-then-show-that-fprime0-exists

Name: sketch the graph of fxxx then show that fprime0 exists
Uploaded: 2019-12-14
Duration: 3 min 50 s
Channel: Numerade
Description: Sketch the graph of $f(x)=x|x|$ . Then show that $f^{\prime}(0)$ exists.

Question

Sketch the graph of $f(x)=x|x|$ . Then show that $f^{\prime}(0)$ exists.

Foster Wisusik · Accepted Answer

Okay, so this right here is the graph of F of axe. Cols X times absolute value backs And it wants us to show that the derivative not only exists at X equals zero, but also wants us to find that value. So how do you show that something is differential? Well, just like limits and continuity, the right derivative has to equal the left derivative. So we&#x27;re gonna end up making this function in the two pieces a right piece in a left piece and show that they have equal derivatives. So how do we do that? Well, we&#x27;d really like to get rid of that absolute value. So we&#x27;ll just consider each case whether X is positive or negative and see what our function turns into. So So let&#x27;s say f of X at positive values is well, the ex stays on change. The axe is a positive number, and the absolute value is also a positive number. So positive x times positive X is X squared x greater than or equal to zero. But then, if we consider the negative case, we get a negative x times the absolute Valley of Negative X, which is acts so negative x Times X, which is negative X squared if X is less than zero. But we could also put a equal to sign here because F. Of zero is still zero in both of these. So now let&#x27;s consider the derivative piece wise. So F Prime of X is just the derivative of each piece. So the derivative of two X squared is positive to X. If X is bigger than zero or the derivative of negative X squared is negative to X effects is less than or equal to zero. So to show that it&#x27;s differential everywhere, all we need to do is show this boundary 0.0. So f prime of zero from the positive direction is two ex evaluated at zero, which is just zero and f prime of zero from the negative direction. This two x negative two ex Sorry, evaluated at zero, which is again zero. So since f prime of zero plus equals f prime of zero negative zero f prime zero exists and it has a value of zero. And again, just to confirm, we can look at our original graph and see that the tangent line at zero is just horizontal. So our intuition matches

Bon Zapata · Answer

So we need to make a graph of this function and use it defined where the limits exist. So I&#x27;m gonna be doing this and decimals feel free to use any sort of graphing calculator. The only tricky step might be getting these restraints up. I&#x27;ll explain how to do it in decimals, but each program is gonna have its own way of doing active. Okay, so the 1st 1 I&#x27;m just going Teoh, assign this a function name. So after backs is equal to one plus sign racks and to do the restraint, I put a curly bracket from then X is less than zero for the next one. So let me go back for a minute. Even though this is all one function just to not confuse the graphing calculator, I&#x27;m gonna make this free separate functions of. So this 1st 1 was f of X. Then the next one will be tree of axe and then a tree, Becks. So let&#x27;s go back. She of axe is equal to co sign of X arm. Strength is that zero is less than or equal to X is less than or equal to pi. And then our last one will be a true backs and this is equal to sign of acts. Assuming that X is greater than I both from looking at this, we need to figure out where the graph is continuous. The one point of interest is right here will receive this. A jump from the second graph to the feared graph up. You can see we cannot draw this entire graph without lifting up the pencil from the page. We have to lift it up right here. So this gives us a dis continuity and the limit does not exist at this one point. At what point is this? It&#x27;s the point where we switch from graft to to graft free and we switch at pie. That&#x27;s the last point of the 2nd 1 and the first point of the fared line. So this graph is continuous, which means that the limit exists everywhere except at high

Patrick Vaughn · Answer

in this problem. Um, we want to scrap. We want a sketch. A graph of a function where f zero equals zero f prime of zero equals zero f Prime of X is greater than zero for all X, less than zero or X greater than zero. So let&#x27;s go ahead and draw axes Your wine. I&#x27;m sorry, your ex and your wife and we know So we know the point zero comma zero because we have, um we know that f of 00 And we also know that our slope is gonna be zero at that point. So there&#x27;s, like, a potential tangent line. Well, actually, it would be a change that line. Um, so then what else do we know? So we know that, um, when X is less than zero, we have positive slope, so it&#x27;s gonna be like going up, and we know that when X is greater than zero slope is also greater than zero. So that means after this point, the slopes gonna be going up. And if the slope, if the if the function is increasing after a 00 then it&#x27;s gonna be above the X axis. And if it&#x27;s increasing before 00 Then we know it&#x27;s below the X axis because it has to come up to meet that 0.0 So from here, you can just pick any function, you know, that satisfies these conditions. So after top my head, like, why equals X cubed? That kind of shape is gonna mimic the behavior that&#x27;s described in this problem. So we have zero slope in the center and then positive slope on the left and right. Get rid of these rings. We got rid of everything, something like this.

Leon Druch · Answer

Okay here we have a photograph of the piecewise defined function F of X one plus one plus sine of X. When X is less than zero. F of X is defined to be co sign of X for X between zero and pi inclusive, and F of X equals sine of X. When X is greater than high. Looking at the graph of F of X, we can see that it is continuous everywhere, except at X equals pi. Uh, So the limit of F of X as X approaches some number A will exist for every number A except high.

Jonathon Brumley · Answer

we&#x27;re gonna graphic function based on some conditions going to draw just kind of be X Y coordinate system here and above that graph, I will draw some shapes to indicate what&#x27;s happening with it. We are told that F of zero equals two, which is simply the 20.0 to jump put right here for the first derivative were told that F Prime of X is positive for all acts, so that simply means we haven&#x27;t increasing function. Doesn&#x27;t matter what the X value it&#x27;s You are so specifically told that if x zero it&#x27;s positive one. So what? We&#x27;ll talk about that the end. But we have an increasing function if X is less than zero, an increasing function if X is greater than zero and a slope of one when X is zero for the second derivative were told that F double prime of X is graven zero. So positive if X is less than zero and the F double prime of X is negative if X is greater than zero. So that&#x27;s telling us that if X is to the left of the Y axis, it&#x27;s con cave down. Since the second derivative is negative, what sorry. Concave up. Since the second derivative is positive and to the right of the why Axis, the second derivative is negative. So Khan came down. Okay, so on the left hand side, we need a say in increasing graph where it&#x27;s con cave up. So that&#x27;s gonna be the right end. Now, keep in mind that the slope is one when X is at zero. So we want to kind of be careful with how quickly we go up. So right at this point right here, we basically have a 45 degree tangents. On the right side of that, we need an increasing graph that is con caved ounce. That&#x27;s the left sign. So the red would represent what the graph would look like.

Leon Druch · Answer

defined in three different parts. Uh function is defined to be one plus X. When x is less than negative one, that&#x27;s the red portion of the graph that you see. The function f of X equals X squared. When uh negative one is less than or equal to ex uh ex in turn is strictly less than one. That&#x27;s the blue portion of the graph that you see. And last but not least when x is greater than or equal to one. Uh F of X is defined to be two minus X. We want to find all values a along the X axis for which the limit exists. And this graph is uh going to greatly help us. You can see um that this uh function is basically connected. There&#x27;s no breaks anywhere in the function except right here. When X is negative one, we have a break into function when X is positive one. Uh You can see the blue portion of the function uh meets nicely uh the green portion of the function. And so the limit as X approaches one from the negative side, the blue side is going to be the same limit as X approaches one from the uh positive side. The green side. But over here, it when uh X is negative one, or specifically when a is equal to negative one. Uh This function uh does not have a limit in other words. Uh The limit of f of X as X approaches negative one does not exist. Um The limits exist for dysfunction for any other value, any other real number value along the X axis, it does not exist. The limit of dysfunction does not exist when X approaches the value of negative one because as X approaches negative one from the negative side to function uh approaches zero. So the limit of F of X as X approaches negative one from the negative side is zero, but the limit of F of X pay attention to the blue portion of the curve. Now the limit of F of X as X approaches negative one from the positive side, uh F of X is approaching the value of one. And so the limit of F of X as X approaches negative one does not exist because the limit of F of X as X approaches negative one from the negative side does not equal the limit of F. Of X as X approaches negative one from the positive side. And you can clearly see that because the red portion of the graph is getting close to zero, think of the Y axis as to function values, whereas the blue portion of the graph is getting close to the value of one

Problem

Determine the values of $x$ at which the function…

Problem 69 Hard Difficulty

Sketch the graph of $f(x)=x|x|$ . Then show that $f^{\prime}(0)$ exists.

Answer

See graph
slope of the tangent line at $x=0$ is zero, which means that $f^{\prime}(0)$ exists and $f^{\prime}(0)=0$

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See graph slope of the tangent line at $x=0$ is zero, which means that $f^{\prime}(0)$ exists and $f^{\prime}(0)=0$

Calculus for AP

Catherine R.

Anna Marie V.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

Jon Rogawski & Ray CannonCalculus for AP

Catherine R.

Anna Marie V.

Kristen K.

Joseph L.

See graph
slope of the tangent line at $x=0$ is zero, which means that $f^{\prime}(0)$ exists and $f^{\prime}(0)=0$

Jon Rogawski & Ray Cannon

Calculus for AP