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Let $ f(x) = x^2 $.

(a) Estimate the values of $ f'(0) $, $ f'(\frac{1}{2}) $, $ f'(1) $, and $ f'(2) $ by using a graphing device to zoom in on the graph of $ f $.

(b) Use symmetry to deduce the values of $ f'(-\frac{1}{2}) $, $ f'(-1) $, and $ f'(-2) $.

(c) Use the results from parts (a) and (b) to guess a formula for $ f'(x) $.

(d) Use the definition of derivative to prove that your guess in part (c) is correct.

a) $f^{\prime}(0)=0 \quad f^{\prime}\left(\frac{1}{2}\right)=1 \quad f^{\prime}(1)=2 \quad f^{\prime}(2)=4$

b) $f^{\prime}\left(-\frac{1}{2}\right)=-1 \quad f^{\prime}(-1)=-2 \quad f^{\prime}(-2)=-4$

c) $f^{\prime}(x)=2 x$

d) $f^{\prime}(x)=2 x$

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Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

So in this problem were given that our function is F of X equals X squared. And were asked to use a graphing function to determine The values of the derivative at zero at one. Sorry, Ed not one at one half one and at two. All right. So, let's look at our graph here from it. Here's the graph park graphing calculator of F of X equals X squared. So why it goes X squared. So derivative zero. Well, look that that zero is a minimum point down here. And the derivative which is the slope of the tangent line, it's going to be horizontal here, isn't it? That's the only way you're gonna draw a tangent line. That doesn't intersect the graph anywhere else become a secret line. So that's going to be zero. Uh One half. Okay. So let's blow this up a little bit. So that one half here, 1/2. I'm here. Right. And mm had derivative. There looks to be pretty much in parallel with Y equals X, doesn't it? Uh huh. So That means that the slope of one. Okay, what about at one? Well, let's see, appear at one. He's here. That slope looks to be a fair amount steeper, doesn't it? Maybe twice as steep. I think that slopes to Okay, What about appear at two. Well, it to, wow, have a good way up here, don't they? For a little bit. That's here. That slopes a lot steeper in it. That's what it looks to me to be about four. Okay, then it says by symmetry to estimate the value of F. Prime at negative one half F. Prime at -1. And then f prime at -2. All right. Bye symmetry. Well, yep, you can see this graph Is over here at -1/2 -1, -2. Right. These are downward slopes equal to the upward slopes right? Just negatives of them. So my symmetry almost say this is negative one, negative two, negative four. Okay then it says to guests of formula for the derivative. Well, what do I see when the X is a half? I get one when X is one. I get to the next is too I get four. It's like I'm multiplying everything by two. So I'm gonna guess this is two x. And it says to use the definition of derivative definition of derivative to determine the formula for the derivative. Well, definitely derivative says the derivative of F prime of X Is the limit as H goes to zero of F of X plus H minus F of X. All over H. All right, well, let's begin to plug everything in. So this is X plus h. Oops. That almost important part here. The limit as h goes to zero of X plus H squared minus X squared all over. H. This is the limit His h goes to zero X plus h quantity squared. Let's multiply that out. That's X squared plus two X. H plus H squared minus X squared over H. All right. Well, I have X squared minus six square. So that's gone. Everything that's left. I can cancel an H. Out of it. Right? How to each one of those terms? That's left. Okay. So that means I have the limit as h goes to zero of two X plus H don't I? That's all I have left. Well, H goes to zero, that term goes to zero. So that means I'm just left with two x as we had guessed. So, there is the formula for the derivative of our F. Of X.

Oklahoma State University