Continuous Function


by golb0016
Tags: continuous, function
golb0016
golb0016 is offline
#1
Nov30-09, 06:45 PM
P: 16
1. The problem statement, all variables and given/known data
find k for the function so it is continuous and differentiable.
x^2-1 x<=1
k(x-1) x>1

3. 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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statdad
statdad is offline
#2
Nov30-09, 06:51 PM
HW Helper
P: 1,344
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)
n!kofeyn
n!kofeyn is offline
#3
Nov30-09, 09:45 PM
P: 538
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.

golb0016
golb0016 is offline
#4
Dec1-09, 10:45 AM
P: 16

Continuous Function


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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