Where is the function $f(x) = |x| + |x+1| + e^x$ not differentiable

Question

While plotting the function mentioned below using a graphing software quickly gives the answer, I am unable to understand how one must solve it. I have studied some properties of modulus functions and would have been able to draw the graph if the $e^x$ was missing.

Find where the function defined on the real line given by $f(x) = |x| + |x+1| + e^x$ is not differentiable.

Could you please let me know how this question must be approached. Moreover, a hint regarding how a question such as the one mentioned above can be solved if $e^x$ was replaced by some other function like $\log(x)$ or $\sin(x)$

Answer 1

Basically, $e^x$ is differentiable everywhere. So this term causes no issues in the differentiability of $f$. The absolute value terms each cause an issue at the points where their arguments are zero. If $e^x$ was replaced by $\sin(x)$ or $\cos(x)$, the same argument just given holds. If you replaced $e^x$ by $\log(x),$ then the same argument holds $\textit{for $x$ in the domain of the logarithm.}$

Answer 2

I would start this problem by breaking up the derivative of $f(x)$ into multiple terms: $$ f'(x) = \dfrac{d}{dx} |x| + \dfrac{d}{dx} |x + 1| + \dfrac{d}{dx} e^x $$ This does a good job of showing that if there are any points where one of the three terms is undefined, $f'(x)$ must also be undefined at that point.

As (it sounds like) you know, $|x|$ and $|x+1|$ cannot be differentiated at $x=0$ and $x=-1$ respectively. So, neither can $f(x)$ at $0$ or $-1$.

$e^x$ is differentiable everywhere so you don't need to worry about it for this problem in particular.

This same reasoning also applies to the function $g(x) = |x| + |x+1| + \sin x$ .

But what about $h(x) = |x| + |x+1| + \log x$ ?

Just ask yourself: where is the logarithm function not differentiable? Answer: for the range of values $x \le 0$ . So $h'(x)$ does not exist for $x \le 0$ and $x = 0$ and $x = -1$ , i.e. just $x \le 0$.

Answer 3

In this sort of context it's well to be aware that $|x|$ is itself not differentiable at $x = 0$. The reason is because on the left of $0$ the derivative is $-1$ and on the right of $0$ it's $1$.

So knowing that, you can see that at $x = -1$ it is also not differentiable, for the same reasons.

But $e^x$ is differentiable everywhere.

Then you can use a result which says that if $f_1$ and $f_2$ and $f_3$ are differentiable on an interval $I$, then their sum is also differentiable on that interval.

So you have that the function is differentiable on $x < -1$, and on $-1 < x < 0$, and on $x > 0$.

But at the points $-1$ and $0$ it is not.

I will leave it up to you to make the argument rigorous.

EDIT: As for $\log (x)$, be aware that on real $x$ it is not defined on $x \le 0$. So it can't be differentiable on that part of the real number line.

As for $\sin x$, that is of course differentiable everywhere.