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Question 11 (1 point) ListenFind the slope for the tangent line to the graph of the function h(x) = (2x 1(Vz) at (1, 1). Write your answer to 1 decimal place.form ...

Question

Question 11 (1 point) ListenFind the slope for the tangent line to the graph of the function h(x) = (2x 1(Vz) at (1, 1). Write your answer to 1 decimal place.form 941 ( V}pdfCierra Jalan PAYR;341()apdr

Question 11 (1 point) Listen Find the slope for the tangent line to the graph of the function h(x) = (2x 1(Vz) at (1, 1). Write your answer to 1 decimal place. form 941 ( V}pdf Cierra Jalan PAYR; 341()apdr



Answers

Find the equation of the line tangent to the function at the given point. $$f(x)=1 / x^{2} \text { at }(1,1)$$

For the given problem we want to find the equation of the line tangent to the function at the given point. So we're gonna have f of X. Yeah equals one over X squid. Um Well we know that one over X squared is X to the negative too. So when we take the derivative we get a -2 X to the -3. If we plug in X equals one, We're not just getting a -2 as our slope. So our line is going to be y equals -2. X must be, We know the .1. 1. Okay, is in question. So when X equals one, let's copy this over here. So we're gonna have x equal to one And why equal to 1? So negative two plus B equals one. We solve for B by adding two on both sides. So B equals three. Since the equals three. We can write this as y. It was a negative two X plus three and that's our final answer.

Mhm. Really Given problem, let's consider Our Function one over root decks. Yeah, That's gonna be where X is equal to one. So you find the slope of the function at the given point. We see the through local linearity when actually goes closer point, for example, We have 1.00 five. So the rise is a negative 0.0025. Okay. And the run is a 0.005. So the slope is a negative one half. So we have white goes negative one half. That's little B. And we use the point 11 to find our B value, which in this case would be 1.5 oh

With the game pong, let's consider um the slope of the tangent line of the function at the given value. We have F. Of X equals 1/2 X. And then we have that this is going to be an X equals a negative one. So in this case where we end up seeing is that The value of our function is a negative 1/2. Um And the slope of our function at this point resuming closely using local linearity, we see that the slope of our function is going to be when we we have a rise of In this case a rise of zero 005 and a run of .01. Yeah, So we have a slope of one half, which makes sense. So that's going to be how we determine our change in line.

In this problem we will cover the tangent line of a function. So we want to find the tangent line equation for this function epithets at .11. And to do so we must first find the slope of the tangent line by using the derivative. So to find the derivative at X equals one. We'll start off by writing f prime of one equals the limit As X approaches one of flex -F of one Over X -1 substituting and for F of X and f of one we will get one over x squared minus one Over X -1. I'm going to do some work off to the side in green so to simplify the top of our fraction, we're going to combine one over X squared with our one and we will get one minus X squared over X squared. And this can be deconstructed into negative one, X squared minus one. All over x squared again. And the job can be rewritten are factored As a negative one X plus one, X minus one over x squared. And because we have an X -1 at the bottom here, that's like saying Multiplied by one over x minus one. And we see that we can cancel out the X -1 from either side. And so We are left with the limit as X approaches one of negative one times X plus one over X squared. And that will give us a slope. Uh we're plugging one for X. It will give us a slope of -2. So now that we found our slope weekend, proceed to the tangent line equation. And in point slope form our equation will be why -F one Equals the derivative at one, multiplied by X -1. This is going to be Y -1 equals negative two times X -1. And when we multiply everything out and move the one from the left to the right, we will get the tangent line equation of Y equals negative two X plus three. Yeah. And lastly we have our graph to the left And green is the graph of F of X. And we have the .11 shown with an arrow. And we have our tangent line and Pink, which is R Y equals negative two X plus three. And we see that the tangent line crosses our graph or touches our graph At the .11. And that is all for this problem.


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