Find k for Continuous & Differentiable Function

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Homework Help Overview

The discussion revolves around finding the value of k for a piecewise function defined as x^2 - 1 for x ≤ 1 and k(x - 1) for x > 1, ensuring that the function is continuous and differentiable at x = 1.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the conditions for continuity and differentiability at the point x = 1, discussing the need to match left and right-hand derivatives and the implications of different values of k.

Discussion Status

Some participants have provided insights on checking continuity and differentiability, suggesting that continuity holds for any k, while differentiability requires matching slopes at x = 1. There is an ongoing exploration of the implications of these conditions.

Contextual Notes

Participants are considering the definitions of continuity and differentiability, particularly at the transition point of the piecewise function, and are questioning the assumptions regarding the behavior of the function at that point.

golb0016
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Homework Statement


find k for the function so it is continuous and differentiable.
x^2-1 x<=1
k(x-1) x>1

The Attempt at a Solution



k(x-1)=0 for x=1
k(0)=0
k = 0/0?

How do I know if the function is differentiable?
 
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Work on making it differentiable at x = 1 first - if you can, then you know it will be continuous there. (what are the left / right-hand derivatives)
 
Actually, that function is continuous for any value of k. The only point you have to worry about, both for continuity and differentiability, is the point x=1. You can explicitly show that the function is continuous at 1 by computing the limit of f as x approaches 1 and show that it equals f(1). You'll see it doesn't matter what k is for continuity.

To worry about differentiability, all you need to check is that the slope of the function k(x-1) at x=1 matches up with the slope of x2-1 at x=1. Remember that for a function to be differentiable at a point, the limit of the difference quotient at that point must exist. By checking that the slopes match up, you are checking that the left and right handed limits of the difference quotient equal each other (i.e., checking that the limit exists). The function is differentiable everywhere else since it is defined there by polynomials.
 
f(x) = x^2-1
f'(1) = 2(1) = 2

f(x) = k(x-1)
f'(1) = k = 2

Therefore k=2 and is differentiable at this point now, correct?
 

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