If possible, find a and b

  • Thread starter Kolika28
  • Start date
  • #1
142
26

Homework Statement:

##g(x)=\left\{
\begin{array}{ll}
\frac{e^{ax+b}-1}{x}, & x>0 \\
\frac{x}{2}+1, & x\leq 0 \\
\end{array}
\right.##

If possible, find the values ##a## and ##b## that make the function g(x) differentiable.

Relevant Equations:

The derivative
So this is what I'm thinking:

After watching some YouTube videos on the subject, the first thing I do is check for continuity. So I plug in for ## x=0## and is left with ##
\frac{e^{b}-1}{0}=1##. I don't think I'm doing this right given the fact that I'm left with 0 in the denominator. Afterwards I was supposed to set the derivative equal of the to expression in the function equal to eachother, but only if it was continuous¨.

But I can't have 0 in the denominator. So does that make the function discontinuous and therefor not differentiable?
 

Answers and Replies

  • #2
13,582
10,710
What "would" be the derivative at x=0x=0? Have you differentiated eax+b−1xeax+b−1x?

Also, you are right, it has to be continuous first. So which combination makes ##\lim_{x \to 0}g(x)=1##?
 
  • #3
RPinPA
Science Advisor
Homework Helper
571
319
the first thing I do is check for continuity.
Good first step. It's not differentiable unless it's continuous. Remember what that means: that the limit as ##x \rightarrow 0## is the same when approaching from the left and the right. Sometimes that's the same as plugging in ##x = 0##, but in general that's not what limit means.

But I can't have 0 in the denominator. So does that make the function discontinuous and therefor not differentiable?
It does if the numerator ##(e^b - 1)## is nonzero, in which case there's a vertical asymptote at ##x = 0##. The limit is either ##+\infty## or ##-\infty##.

So the only hope for continuity is if ##(e^b - 1)## is 0 at ##x = 0##. In that case, plugging in 0 gives you an indeterminate form ##0/0## and that is NOT the limit as ##x \rightarrow 0^-##. You want to evaluate that limit and see if it can be made the same as the limit as ##x \rightarrow 0^+##.

Provided the function can be made continuous, you aren't done. Continuity is necessary for differentiability, but it isn't sufficient.
 
  • Like
Likes Kolika28
  • #4
142
26
What "would" be the derivative at x=0x=0? Have you differentiated eax+b−1xeax+b−1x?

Also, you are right, it has to be continuous first. So which combination makes ##\lim_{x \to 0}g(x)=1##?
So should I in this case differentiate first?
 
  • #5
13,582
10,710
You should make sure that there is no gap first. Then as @RPinPA has said, you are not done, yet. The derivative has to exist. So it's reasonable to look what the derivative looks like. It isn't allowed to have a gap either, if it is differentiable at ##x=0##.
 
  • Like
Likes Kolika28
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
956
The derivative of a differentiable function is not necessarily continuous but it does have the "intermediate value theorem" (that is what is meant by "no gap"). So if you find the derivative for x> 0 and the derivative for x< 0 and take the limit of each as x goes to 0, either the two limits are the same, and the derivative of the function is that value, or the two limits are not the same and the function is not differentiable at x= 0.
 
  • Like
Likes Kolika28
  • #7
142
26
Ok, I not sure if I'm understanding correctly. But maye it's best if I try to show you what I'm doing with the information given, and then take it from there:

So I check for continuity to make sure that the function does not have a gap.
##\lim_{x\to 0^{-}}\frac{x}{2}+1=1##
Then we must have that
##\lim_{x\to 0^{+}}\frac{e^{ax+b}-1}{x}=1##

But how do I make sure if this gap does exist or not?
 
Last edited:
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Yes, that limit "must be 1". Can you determine whether it is or not? I see, for example, that if b is not 0, the limit does not exist at all because the denominator is going to 0 while the numerator is not. In order to have a chance of having a limit, the numerator, [itex]e^{ax+ b}- 1[/itex] must be 0 for x= 0. That, again, is true if b= 0. Now, since [itex]\frac{e^{ax}-1}{x}[/itex] is the "indeterminate" form, [itex]\frac{0}{0}[/itex], at x= 0, you can use L'Hopital's rule.
 
  • Like
Likes Kolika28
  • #9
142
26
I finally got it! Thank you so much both of you!
 

Related Threads on If possible, find a and b

Replies
3
Views
1K
Replies
11
Views
790
Replies
2
Views
486
Replies
2
Views
2K
  • Last Post
Replies
6
Views
7K
Replies
8
Views
3K
  • Last Post
Replies
2
Views
757
Replies
3
Views
3K
  • Last Post
Replies
8
Views
872
Top