Can a Function be Differentiable if it is Not Continuous?

[Kolika28]

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 each other, 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?

 

[fresh_42]

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 $\lim_{x \to 0}g(x)=1$?

 

[RPinPA]

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.

 

[Kolika28]

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?

 

[fresh_42]

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 $x=0$.

 

[HallsofIvy]

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.

 

[Kolika28]

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?

 

[HallsofIvy]

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, e^{ax+ b}- 1 must be 0 for x= 0. That, again, is true if b= 0. Now, since \frac{e^{ax}-1}{x} is the "indeterminate" form, \frac{0}{0}, at x= 0, you can use L'Hopital's rule.

 

[Kolika28]

I finally got it! Thank you so much both of you!