Question

(a) For what values of $x$ is the function $f(x)=\left|x^2-9\right|$ differentiable? Find a formula for $f^{\prime}$. (b) Sketch the graphs of $f$ and $f^{\prime}$.

   (a) For what values of $x$ is the function $f(x)=\left|x^2-9\right|$ differentiable? Find a formula for $f^{\prime}$.
(b) Sketch the graphs of $f$ and $f^{\prime}$.
Single Variable Calculus: Early Transcendentals
Single Variable Calculus: Early Transcendentals
James Stewart,… 9th Edition
Chapter 3, Problem 77 ↓
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(a) For what values of $x$ is the function $f(x)=\left|x^2-9\right|$ differentiable? Find a formula for $f^{\prime}$. (b) Sketch the graphs of $f$ and $f^{\prime}$.
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Transcript

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00:03 All right.
00:04 So we have function absolute value of x squared minus nine.
00:08 And we gotta find where is this function? defensible and find the derivative and then part b graft.
00:15 Both those.
00:16 So we're going to start with what the absolute value function actually represents.
00:23 It's a combination of functions.
00:26 So you have one expenses greater than three.
00:31 You know, x squared minus nine.
00:36 When exit between three and negative three, you're gonna have x squared minus nine.
00:46 Negative.
00:48 And when x is less than negative three ganyu off x squared minus nine.
00:55 All right, then, using the definition of the limit, we have the our definition of the derivative.
01:07 You have purchase a f of x minus f of a over x minus.
01:19 Hey, now, just looking at the the f of x function that we have on the left, you can tell that that function changes at two points, it changes at three.
01:30 And edges and negative three, so we can anticipate derivatives.
01:36 Uh, not being, ah not having values at those points because of the function.
01:42 Changes on the derivative is gonna change.
01:47 So we'll test the derivative at x equals three first.
01:58 So the limit x approaches three like after 30 and then we plug in our function to get all right.
02:33 So in order for this limit to exist, the limit has to exist coming from the left and the right.
02:41 So we'll continue that over here with from the left when it's coming from the left.
02:52 We're gonna use this middle function because left means it's less than three.
03:07 Factoring dividing gives you expose three negative expects three and then plugging in three.
03:20 Getting out negative.
03:21 Six.
03:22 No, they're the same thing coming from the right.
03:30 Coming from the writing music x is gonna be greater than three.
03:32 So we use the top function x squared minus nine.
03:40 It's gonna come down to expose three plugging in three years uni at six.
03:45 So as you can see, these two are not the same.
03:52 So therefore the limit.
03:57 His ex approaches three x squared minus nine.
04:05 It does not exist, and therefore, if this doesn't exist, it means it's not differential at three.
04:14 And that was for equals.
04:16 Three.
04:17 Now we'll examine a equals negative three approaching from the left.
04:36 That means it's less a negative three system with that first function or 1/3 function right there.
04:46 Watch the sign right here because you're doing x minus a.
04:49 So it's gonna be positive.
04:50 Three plugging in negative three.
04:57 Gonna get negative.
04:59 Six.
05:07 Right.
05:07 So from the right means it's greater than *** three.
05:09 It's gonna be the middle function right there.
05:22 It's a negative.
05:24 X minus three.
05:26 Plug in bigger three.
05:28 You're going to get positive...
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