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(15)Let fl)= (a) Find f'(~) e"+-' when x = -1 Find the equation of the tangent line to f (x)= (b)...

Question

(15)Let fl)= (a) Find f'(~) e"+-' when x = -1 Find the equation of the tangent line to f (x)= (b)

(15) Let fl)= (a) Find f'(~) e"+-' when x = -1 Find the equation of the tangent line to f (x)= (b)



Answers

In Fig. 19, find the equation of the tangent line to $f(x)$ at the point $A.$

This time they ask, Find the tension point. Let's see one comma zero That's the calculations. Now the life value one times and ensure a longer one. I'm also gonna need a slope. So let's take our derivative. We're going to need to do product Well adapted to the X Natural Aga Rex What was timely that went on Well, uh, one of ranks, some of the nominee it cancels out there. I ended with X with one plus steer natural Lagerback. If we want to find our derivative about one, we could just plug it in only to end up with one has our value. Because isn't it with a lot of one, which is a zero there? Ah, let's find the equation of our attention line, and then we end up with one times X minus one. So we have why is equal to X the tangent line

Give us the function. F X is equal experiment times, the natural other banks. Well, let's see. They want us to find the equation of the tangent line, that one. Let's find the Y value. We'll get one squared times the natural log of one Alexis one time zero So we just get zero. Let's try to find the slope so we'll take the derivative using product role. The first factors. Derivatives two x leave the natural of alone. Then we x squared alone. Take the derivative of the natural long when we end up with Let's just in X ray here and we can factor that out to get X one plus two natural on do X. So if we tried to find the slope at one, that would just speak one plus two times the natural log of one. But that's just to time. Zero. So we just end up with one is our slope. It's a point. So for him, why minus zero is equal to the slow one times X minus one and we end up with our tangent line being Why is equal to X minus one. Me

We are wanting to write the equation for the line that is tangent to equals f of X equals three X squared over five X squared plus seven X. At the x value of one. Well, to write the equation for a tangent line, the first thing we need to determine is the slope of the tangent line, slope is going to come from the derivative. The derivative of our function will fit the quotient rule. Now, before I do use the quotient rule. Since there is a common acts, let's go ahead and reduce that out. So that's three X over five X plus seven. Using the caution role, we have five X plus seven Times The derivative of the top three -3 x times the derivative of the bottom five. All over our denominators square. We can simplify that down. That would be 15 x plus seven. Show me plus 21 3 times seven minus 15 X. All over five x plus seven squared and with the 15 X is canceled canceled. We have 21 over quantity five X plus seven quantity square. We are looking for the equation for the tangent line when X is one. So if we fill one in, What we will find is that why prime is 21 over 12 squared? 21 Over 144, Which will reduce to 7:48s. So using Y equals mx plus B. With a slope of 7 48. We currently have an equation of white will 7 48 X plus B. And to determine what B is. We need to know what point on the graph we're talking about. Well, if X is one if we fill one in for X, what we're going to find is that why is 3 12 or 1/4? So we need to fill in the point one 1/4. If we do so, will have 1/4 equals seven 48 Plus B. Subtracting the seven 48th, 1/4 minus 7 48. And really, that's 12 48 minus 7 48 or five 48th. So if we fill that in for be, our final equation is why equals 748 x plus 5 40 eight's.

Here were given that ffx is equal to X squared. Now, to find an equation of the tangent line to the graph, we first find the slope of the curve by differentiating our equation as we find the derivative which will give us the slope off the tangent line. So therefore we have our function effects equals x squared. So then the derivative that would be half prime of X. Okay, well, that's equal to the limit. As a church approaches zero of, um well, f of X plus h square eso well f of X plus age, which is X plus a h squared and then minus the functions of minus FX. So minus oops minus um X squared, all divided by h. Okay, so we go through this and we get that we have to we will get X squared. Plus two h x plus a squared minus X squared all over H s O. The X squared minus X squared cancels out. We get to h X plus a squared all over h we factor out in h um, so we just get the limit as H approaches zero off two x plus h, which is equal to two X, right? So that's by the definition. Or we could just use the power role here, right? It just brings down the two. Andre, get right away. We get two X. Okay, So the derivative of our function is equal to two X now for part A, um, since the derivative. So since F Prime of X is equal to two X, then the slope of the tangent line, um at X is equal to three. Well, that's just out prime off three, which is just equal to two times three, which is equal to six. So the equation of the tangent line, um, to the graph at the point, while when x is three right f of X is six. So at the point, um, well, three squared is nine. So at the 90.3 comma, nine eyes six. Right, that the slope of the tangent line when excess three, when x is three, plugging it into our function x squared while the FX is nine. So the slope of the tangent line at the 90.3 common nine is six. So therefore we can use our point slope form of a line which the point slope form member is well, we have y minus y one is equal to the slope. M times X minus X one, right for some given point X one comma Y one. So we have a point and we have the slope, right? If we have the point, we have any point, um, on our function, we have the swell for a line, but we have any point on the line and the slope of the line. We can then make the equation of the line by using this form right here. So again, this is a slope of the tangent line to the graph. So we have why minus y one, what are Y value here is nine. So why minus nine is equal to the slope six times X minus the X value, which is three. So eso Here's we have y minus. Why minus nine is equal to six x minus three. If we distribute and solve for why we get that Y is equal to six X minus nine. Mhm. All right, so there is Theo equation of a tangent line. Okay. And then for part B, well, again we have our derivative is of prime affects um is equal to X. So now we have. We want the slope of the tangent line when X is equal to negative one. So f prime of negative one. Well, that's just equal to two times negative one, which is negative. Two. So then we used the equation off the tangent line. So again we have Why, minus y wants to hear the y minus y one y value is one y minus. One is equal to the slope, which is negative two times x minus X one X minus and negative one, which becomes X plus one and again we software. Why here? And we get that Why is equal to negative two X minus one. All right, and then for C. We want the slope of the tangent line when actually equal to 10. So actually go to 10 or what? Um, what's the What's the slope of the tangent line? Well, he's plug it into our derivative formula, so we do f prime off 10. That's equal to just two times 10, which is equal to 20. So then, using our point slope form of the line, we have y minus y one or y minus 100 is equal to the slope 20 times X minus X one So X minus 10 and again solving for why we get that Why is equal to well, just distribute 20 and then add over 100. I mean, y is equal to 20 X, um minus are yes, 20 X um, my 100. And there we have it.


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