Problem 64 Hard Difficulty

The left-hand and right-hand derivatives of $f$ at $a$ are defined by
$$\begin{array}{c}{f_{-}^{\prime}(a)=\lim _{h \rightarrow 0^{-}} \frac{f(a+h)-f(a)}{h}} \\ {\text { and } \quad f_{+}^{\prime}(a)=\lim _{h \rightarrow 0^{-}} \frac{f(a+h)-f(a)}{h}}\end{array}$$
if these limits exist. Then $f^{\prime}(a)$ exists if and only if these one-sided derivatives exist and are equal. (a) Find $f^{\prime}-(4)$ and $f^{\prime}+(4)$ for the function
$$f(x)=\left\{\begin{array}{l}{0} \\ {5-x} \\ {\frac{1}{5-x}}\end{array}\right.$$
if $x \leq 0$
if $0< x<4 $
if $x \geqslant 4$
(b) Sketch the graph of $f$ .
(c) Where is $f$ discontinuous?
(d) Where is $f$ not differentiable?

Answer

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Video Transcript

and there. 2.8 Number 64 made look pretty intimidating at first, so let's solve it here Together, we're dealing with a piece wise function f so find like this and three separate pieces. And it's going to be very important as we solve this problem, to pay close attention to which of these three pieces are inputs are in. So let's keep referring back often to this definition of our function. F very important. For now, though, let's start with art, Eh? Which asks us to find kind of a funny looking symbol, which we've probably never seen before in our lives because it's just been to find in this problem it's the left hand derivative of the function f at the X value for And the only thing we can do is use the definition that they just gave us. So that's what we will. D'oh! We're just going to copy the definition. They just told us in this problem, limit his H approaches zero from the left off big fraction F of Now you noticed that the a value in our cases for so we can simply substitute the number four anytime we see the variable A in their definition, So let's do that right. Four plus age, minus F of once again instead of a he will write four all over age. Now, here's where it's very important to notice. As H approaches zero from the left of this means that H is going to be less than zero. This is important because looking at our input here for plus age, we need to figure out in order to write out what f will be, you need to figure out which part of the domain four plus ages in which of these three pieces. So let's look at H is less than zero. This means four plus age is less than four. Make sure you're convinced a vet before we move on, and therefore, since our input right here is less than four, well, let's look at puts us in this second piece of the domain. Whatever input is less than four. That means the function is defined as five minus the input. So let's go back, and we will use that as we continue. So this part's still the same. We just said half of four plus age is really five, minus the input. The input is four plus age. Now we have to be careful again f of four to figure out what that is. Once again, let's look back at the function. We want half of four, and we can look well. Four, certainly in this part of the domain, since for is greater than or equal to four. Therefore, will use that part of the definition with the input for which is one over one, which is one so effort for is one. Let's remember that. What? And me right One instead of f before. Okay, this looks not too bad. Now, uh, you will keep this part five minus four minus age minus one. Just a little arithmetic on top will give us five minus four is one minus one is zero were left just with this in his H approaches. Zero from either direction. Certainly from the left. This should be a pretty familiar limit. We're going to get negative one. And that's the answer for the first part of part A, the left hand derivative of F at four is minus one. Okay, we're halfway through part name now to get the right hand derivative, we simply do the exact same thing using their definition, and only one small thing will change. So let's take and right hand derivative of F at 43 Only thing that changes if we look at their definition is that a JJ well, now approach zero from the right. Everything else is the same once again. F four plus h minus fo four all over age. Once again, it's important that H is approaching zero from the right. This means eight will be greater than zero, and that means four plus age will be greater than four. Looking back at our function F if our input is greater than four, don't that means we're in this part of the domain. So our definition is an F of some input. Is one over five minus that input? Let's remember that as we carry on. So we just said EFF of our input. It's one over five, minus that input in our input was four plus h minus f f. Or we can remember from last time f f or is just the number one that hasn't changed all over it. And now just a little clean up as often with calculus problems There's a little bit of calculus in a lot of algebra, so just make sure that we don't make any mistakes here in computing. These little fractions five minus four is one minus. H There. We're going to need to combine these two fractions on top in order to simplify. So let's do that now. Let's rewrite one as one minus age over one minus H. That's where the two denominators air the same and will be able to combine the next step. We're almost there now. The denominators are the same, and we can simply we know that the nominators will be one minus h subtracting across the top one minus one. And this would be plus age. Be careful. That's a common mistake. We're subtracting all of this, so it's minus one minus minus. H gives us a plus age. All right, nearly there, one minus one plus h. It's It's just a tch on top one, minus h over H and coming to the end here. You can rewrite these in many ways, perhaps easiest to see if we just we write like this. And once again, since H is never equal to zero, were simply approaching zero. Those ages will cancel, and hopefully we can see and that we're left with limit is H approaches zero from the right one over one minus h, which is one. And we have finished part. Kate, that is our right hand derivative. At four. It's just one part. Be asked us to graph the function F soap to graph it. Let's just look back at how f is defined. Let's first graph this first section. So whenever X is less than equal to zero, the function is simply zero, so that's be not too bad. So any Time X is less than or equal to zero. So in this part of the graph, our function is simply zero. We know how to graft that simply a straight line with the height of zero. Any time we're less than equal to zero Good. The next part when exes between zero and four, our function is five minus X. So we just said between zero and four that this part right here five minus x Well, that's just a straight line with a slope Negative. One hand. Oh, I intercept five. So we know how to sketch that, and we stop once we hit X equals four. That's open circles on either end there, and that takes us to our last over here, our last part of the definition, when X is greater than or equal to four won over five minus X. Okay, so when X is greater than or equal to four, we'll make that blue. So this whole part of the graph one over five minus X, and we hopefully recognize that as well. You can see there's an S in total positive five so you can sketch that in the dotted line is usually see them and will assume you're somewhat familiar with this graph. Here we can tell as X approaches five fromthe left. We're going to have a limited positive infinity, so it will go up forever as we could approach five from the left to figure exactly the height of four. Let's just plug in. Four for X five, minus four is one, and we already knew from doing the problem. F of four equals one. Don't forget effort for equals one. So every go effort four equals one. That's just connect these nicely. It's possible. And then on the other side, something along these lines here. You noticed that as we approach five fromthe rights, it won over five Minus X will have negative infinity in these rural all grafts that were familiar with from earlier in the text. So there's our graph and that completes part B parts C and D. Well, let's use a graft to answer them. You might as well use our hard work. Where is F discontinuous were after in part. See? Well, we can simply look and shouldn't be too hard. We can tell discontinuous the only places certainly at X equals zero. We have ah, jumped is continuity. You can see right here from the left. We're going toward a height of zero from the right toward a height of five. So X equals zero, uh, four is Okay, we do have Ah, continuous is continuous at X equals four from the left and from the right, we're approaching the same height, which, uh is also effort form and the vertical ask. Himto then, is the only other dis continuity. So X equals five knows are the two X values that which our function is disc continuous. Those are the only ones and finally part deep Where is F not differential? Well, we can see most of the graph. Certainly is differential when X equals zero less, very less than or equal to zero. Our function is simply f of X equals zero that is defensible on, as is during zero everywhere here. Certainly, since you cannot be differential, function is not differential. If it's not even continuous X equals zero must be a place in the domain where the function is not differential. This entire part here between zero and four is differential and we can even see the derivative is negative. One at four. Remember they told us that F prime of four the derivative exists on Lee. If he left and right and derivatives equal each other, let's just look at our work from part. Eh? The left hand derivative at four was negative one. The right hand derivative at four was positive. One. They're not the same. So clearly f prime of four must not exist. You can see that just by looking here as well. The function comes to appoint a cusp, so there's no derivative at four and likewise X equals five, not even continuous there, so it cannot be differential. Those were the only three places where our function is not differential. So a bit of a whirlwind of there. I know there were a lot of steps, but hopefully this made sense helpfully, This was helpful.