a. Let f(x) be a function satisfying |f(x)| ≤x^2 for -1 ≤x ≤1 Show that f is differentiable at x

A. Let f(x) be a function satisfying |f(x)| ≤x^2 for -1 ≤x...

Key Concepts

Piecewise Functions and Continuity/Differentiability

Piecewise functions require careful consideration at the points where their definitions change. To establish differentiability at such points, it is necessary to check that the limit of the difference quotient exists and is the same from both directions. This ensures that the function transitions smoothly between the different pieces.

Behavior of Functions Near a Point

Understanding how functions behave in the vicinity of a point, especially in terms of their growth or oscillatory characteristics, is essential in assessing continuity and differentiability. For instance, a function that is bounded by a quadratic function near zero will have its oscillations dampened sufficiently, ensuring that the derivative exists at that point.

Squeeze Theorem

The Squeeze Theorem is a method used to determine the limit of a function when it is bounded between two other functions that converge to the same limit at a particular point. This theorem is particularly useful when dealing with functions that are difficult to evaluate directly, as it allows one to 'squeeze' the target function between bounds that simplify the computation of the limit.

Limit Definition of the Derivative

The derivative of a function at a point is defined as the limit of the difference quotient as the increment approaches zero. Specifically, f'(0) is given by the limit as h approaches 0 of [f(h) - f(0)]/h. This concept is crucial in testing whether a function is differentiable at a point.

Transcript

00:01 In this problem, in part a, we're said, we're given, let f of x be a function satisfying the absolute value of f of x is less than or equal to x squared for all x between negative 1 and 1.

00:20 And we're asked to prove that f is differentiable at 0, and then we're asked to find f prime of 0.

00:26 So let's just write down the limit that we know.

00:28 So the limit is h goes to 0 of f of x plus h minus f of x.

00:35 X over h.

00:39 Oh, sorry, this should be f of zero plus h, since we're interested in x equals zero.

00:45 This is f of zero plus h minus f of zero.

00:49 So again, the question is, what is f prime of zero? so we need to kind of figure out what all these terms are.

01:00 So for the f of zero term, we aren't told what f of zero is explicitly, but we can still figure it out.

01:07 We know that the absolute value of f of zero is less than or equal to zero squared because this is the condition that's given to us.

01:14 This is this condition here, which is just zero.

01:20 So if the absolute value of f of zero is less than or equal to zero, this tells us that f of zero has to be equal to zero because there's no numbers other than zero whose absolute value is less than or equal to zero.

01:31 So the zero is nice.

01:32 That's good.

01:33 So let's rewrite this.

01:34 This is a limit as h goes to zero of f of h over h.

01:40 Now again, we got to kind of understand what f of h looks like.

01:44 So once again, let's use this condition.

01:49 So we know that f of h in absolute value is less than equal to h squared for negative 1 less than equal to h less than equal to 1.

01:58 So in particular, we only care about what's happening when age goes to zero.

02:01 So we only care about very small values of h.

02:04 So we can rewrite this as negative h squared less than or equal to f of h less than or equal to h squared.

02:11 So we get rid of these absolute value signs.

02:14 Now let's divide everything by h...