Find f^'(c) if it exists. f(x)={ x+1, x ≤-1 ...

Key Concepts

Piecewise Functions

Piecewise functions are defined by different expressions over different intervals of the domain. Analyzing such functions requires careful attention at the points where the definition changes, as these are the locations where the behavior of the function may differ, leading to potential discontinuities or non-differentiability.

One-Sided Derivatives

One-sided derivatives involve computing the derivative from the left (left-hand derivative) and the right (right-hand derivative) separately. At a point where the function's definition changes, these values must be compared; if they are equal, the function is differentiable at that point, and if not, the derivative does not exist.

Differentiability at a Point

Differentiability at a specific point requires that the limit of the difference quotient exists and is the same from both sides. For piecewise functions, checking differentiability involves computing the left-hand and right-hand derivatives at the boundary and confirming that they are equal.

Transcript

00:03 F of x is a piecewise defined function.

00:08 It equals this expression when x is less than or equal to negative 1.

00:13 And f of x is equal to this expression when x is greater than negative 1.

00:19 We need to find f prime of negative 1 by taking the limit as each approach is 0 of the difference quotient.

00:28 F of negative 1 here, let's go ahead and find that real quickly.

00:32 We'll write that down here.

00:34 Of negative 1.

00:39 Okay, what is our function value when x is negative 1? well, if x equals negative 1, that fits in this inequality here, which means we're going to use this expression for f of x to calculate f of negative 1.

00:55 So f of x is x plus 1, so f of negative 1 is negative 1 plus 1, which is 0.

01:05 Now here's where we're going to run into a little bit of trouble.

01:09 We're trying to find the first derivative at negative one.

01:12 Notice that the function is broken up into two different expressions right at negative one.

01:20 So when we find a limit of the difference quotient as h approach is zero, we run into a little bit of trouble writing down f of negative 1 plus h.

01:32 And here's why.

01:33 When we take the limit as h approach zero, it has to approach from the right side, of zero and from the left side of zero.

01:40 If h is approaching zero from the right side, that means it's a tiny positive number, and negative 1 plus h will be negative 1 plus a tiny positive number, which makes it a little bit greater than negative 1.

01:53 So we need to use this expression for f of x when h is approaching zero from the right side.

01:59 But if h is approaching zero from the left side, the negative side, then h is a tiny negative number, negative 1 plus a tiny negative number will be something less than negative 1.

02:12 So when h approaches 0 from the left side, negative 1 plus h is less than negative 1.

02:20 And so we have to use this expression to calculate f of negative 1 plus h when h is approaching 0 from the left side.

02:29 So what we're going to have to do to find the limit of this difference quotient as h approaches 0 is we're going to find a limit as h approaches 0 from the positive side.

02:41 We'll do it down here.

02:42 And we're going to have to find the limit of this difference quotient as h approaches zero from the negative side, the left side.

02:50 We'll do it down here.

02:58 Okay, so we're getting ready to take the limit of the difference quotient as h approaches zero from the positive side.

03:06 So first we need to write down what is f of negative 1 plus h.

03:10 If h is approaching 0 from the positive side, that means it's a tiny positive number...

Please give Ace some feedback