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a. Let $f(x)$ be a function satisfying $|f(x)| \leq x^{2}$ for $-1 \leq x \leq 1$Show that $f$ is differentiable at $x=0$ and find $f^{\prime}(0)$b. Show that$$f(x)=\left\{\begin{array}{ll}{x^{2} \sin \frac{1}{x},} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right.$$is differentiable at $x=0$ and find $f^{\prime}(0)$

(a) $f^{\prime}(0)$ exists and equals $0 .$ See inside for the detailed explanation.

Calculus 1 / AB

Chapter 3

Derivatives

Section 2

The Derivative as a Function

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in this problem in part A. We're said we're given what f of x b a function satisfying the absolute value that of FX is less than or equal X squared for all ex Betweennegative one and one and were asked to prove that as his defensible it zero. And then we're asked to find f prime of zero. So it's just right down the limit that we know. So the limit is h goes to zero of f of x plus h minus FX over age. Oh, sorry. This should be f zero plus agents were interested in X equals zero. This is F zero plus H minus zero. So again, the question is what is f primavera? So we need to kind of figure out what all these terms are. So for the F zero term, we aren't told what death of euro is explicitly, but we can still figure it out. We know that the absolute value of F zero is less than or equal to zero squared because this is the condition that's given to us. This is his condition here, which is just zero. So the absolute value of zero is less than or equal to zero. This tells us that F zero has to be equal to zero because there's no numbers other than Jiro whose absolute value is less than or equal to zero. So the zero is is nice. That's good flattery, right? This is a limited H goes zero of half of age over eight. Now get in. We've got a kind of understand what FF H looks like. So once again, let's use this condition. So we know that f of H in absolute value is less from ableto age squared for negative, one less cynical age last Niccolo won. So in particular, we only care about what's happening with age. It goes to zero, so we only care about very small values of age. So we can rewrite this as negative age where, less than or equal to f of H less than or equal to H Square. So we get rid of these absolute value signs. Now it's divide everything by H. So we have h negative age less than or equal to half of H over h less than or equal to H. So once we have this, we can finally sort of plug in plug into our women. Well, I really plug in, but we can use the squeeze The're, um for limits or the sandwich theorem for limits. So we know that the limited H goes to zero of H is equal to zero, and the limit is h ghost. Zero of minus age is also equal to zero and f f h over h is squeezed between or sandwich between negative agent H. So this tells us that the limit is Asia goes to zero of death of H over A. JJ must be equal to zero by the sandwich there. Great. Now, in part B, you were asked to consider specifically the function f of X equals X squared times sine of one over X x is not equal to zero and zero if X equals zero and were asked to calculate f prime of zero. This is kind of hard to just to do, but luckily we just did part a. So let's show that that FX satisfies the assumption from part eh, namely, we just need to check if the absolute value of FX is less than or equal to x squared. If we can show this, then then we'LL be done. So. Is this just for negative one less stable excess, Michael The one. So let's calculated the absolute value. Beth of X is equal. The absolute value of X square times sine of one over x chaps, the value of X squared times the absolute value of sign of one over X. But we know how sign works. We know that the absolute value of sign is always less or equal to one, because sign only goes between minus one and one. So we're just left with the absolute value of X squared times one but X is always positive or seen. The X squared is always positive. So the absolute value that's where it is just X squared. So we do see that Oh, this should have been less than or equal to once we plugged in. Our minds are one there she have to value. SFX is less than or equal to x squared. So dysfunction satisfies the hypothesis from part A. So if we apply part, eh? We know that f prime of zero exists and is equal to zero

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