MHB Find f'(0) from a piecewise function

[Dethrone]

Find $f'(0)$ if:
$$f(x)=\begin{cases}\frac{g(x)}{x}, & x\ne 0 \\[3pt] 0, & x=0 \\ \end{cases}$$
where $g(0)=g'(0)=0$, $g''(x)=17$ and $g''(x)$ is continuous at $0$.

Progress:
First,

$$f'(x)=\begin{cases}\frac{xg'(x)-g(x)}{x^2}, & x\ne 0 \\[3pt] 0, & x=0 \\ \end{cases}$$

Wouldn't $f'(0)$ simply be 0, from above, or is there a flaw in my logic? I mean, I could probably prove that the function is continuous at 0 from the left and right side by l'hopital's rule...tell me if I'm right so far.

EDIT: I think I determined that $\lim_{{x}\to{0^{+/-}}}\frac{xg'(x)-g(x)}{x^2}=\frac{17}{2}$, but why is it not equal to $0$?

 

[Euge]

Find $f'(0)$ if:
$$f(x)=\begin{cases}\frac{g(x)}{x}, & x\ne 0 \\[3pt] 0, & x=0 \\ \end{cases}$$
where $g(0)=g'(0)=0$, $g''(x)=17$ and $g''(x)$ is continuous at $0$.

Progress:
First,

$$f'(x)=\begin{cases}\frac{xg'(x)-g(x)}{x^2}, & x\ne 0 \\[3pt] 0, & x=0 \\ \end{cases}$$

Wouldn't $f'(0)$ simply be 0, from above, or is there a flaw in my logic? I mean, I could probably prove that the function is continuous at 0 from the left and right side by l'hopital's rule...tell me if I'm right so far.

EDIT: I think I determined that $\lim_{{x}\to{0^{+/-}}}\frac{xg'(x)-g(x)}{x^2}=\frac{17}{2}$, but why is it not equal to $0$?

Hi Rido12,

I can see where you're having trouble. Let's start from the top. To find $f'(0)$, use the limit definition

$\displaystyle f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x}$.

Since $f(0) = 0$ and $f(x) = \frac{g(x)}{x}$ for $x \neq 0$,

$\displaystyle f'(0) = \lim_{x \to 0} \frac{g(x)}{x^2}$.

This limit has indeterminate form $\frac{0}{0}$ since $g(0) = 0$. To proceed, use L'hospital's rule (as you've suggested) to get

$\displaystyle f'(0) = \lim_{x \to 0} \frac{g'(x)}{2x}$.

Since $g'(0) = 0$, the above limit has indeterminate form $\frac{0}{0}$. Apply L'hospital's rule once more to obtain

$\displaystyle f'(0) = \lim_{x \to 0} \frac{g''(x)}{2} = \frac{17}{2}$.

The last equality follows from continuity of $g''$ at 0 with $g''(0) = 17$.

Note that it wasn't necessary to compute $f'(x)$ for $x \neq 0$ to get $f'(0)$.

 

[RLBrown]

, but why is it not equal to $0$?

Consider the special case g(x) = (17/2)x²
When x>0
Then
f(x) = (17/2)x
and
f'(x) = 17/2

 

[Dethrone]

Hi Rido12,

I can see where you're having trouble. Let's start from the top. To find $f'(0)$, use the limit definition

$\displaystyle f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x}$.

Since $f(0) = 0$ and $f(x) = \frac{g(x)}{x}$ for $x \neq 0$,

$\displaystyle f'(0) = \lim_{x \to 0} \frac{g(x)}{x^2}$.

This limit has indeterminate form $\frac{0}{0}$ since $g(0) = 0$. To proceed, use L'hospital's rule (as you've suggested) to get

$\displaystyle f'(0) = \lim_{x \to 0} \frac{g'(x)}{2x}$.

Since $g'(0) = 0$, the above limit has indeterminate form $\frac{0}{0}$. Apply L'hospital's rule once more to obtain

$\displaystyle f'(0) = \lim_{x \to 0} \frac{g''(x)}{2} = \frac{17}{2}$.

The last equality follows from continuity of $g''$ at 0 with $g''(0) = 17$.

Note that it wasn't necessary to compute $f'(x)$ for $x \neq 0$ to get $f'(0)$.

Hi Euge! (Wave)

I can follow all your steps, and that at the end, $\displaystyle f'(0) = \lim_{x \to 0} \frac{g''(x)}{2}$ is equal to $\frac{17}{2}$ because $g''(x)$ is continuous at $0$.

However, when I differentiated the piece-wise function $f$ at the beginning, I got:

$f'(x)=\begin{cases}\frac{xg'(x)-g(x)}{x^2}, & x\ne 0 \\[3pt] 0, & x=0 \\ \end{cases}$

Spoiler

The question gives us $f(0)=0$ when $x=0$, so it follows that $f'(0)=0$ when $x=0$ because the derivative of $0$ is $0$. I have also determined that $f$ is continuous.

In the above, when $x=0$, then $f'(0)=0$. What's wrong with that conclusion?

 

[I like Serena]

Spoiler

The question gives us $f(0)=0$ when $x=0$, so it follows that $f'(0)=0$ when $x=0$ because the derivative of $0$ is $0$. I have also determined that $f$ is continuous.

In the above, when $x=0$, then $f'(0)=0$. What's wrong with that conclusion?

Let $h(x)=x$.
Since we have $h(0)=0$ when $x=0$, does it follow that $h'(0)=0$? (Wondering)

 

[Dethrone]

That's true, but I'm getting tangled up here.

Isn't the way I'm getting $f'(x)$ analogous to what you would do with an absolute value function?

$$f(x)=|x|=\begin{cases}x, & x>0 \\[3pt] -x, & x<0 \\ \end{cases}$$

Differentiating both parts of the piecewise function:

$$f'(x)=\begin{cases}1, & x>0 \\[3pt] -1, & x<0 \\ \end{cases}$$

Differentiating yet again:

$$f''(x)=\begin{cases}0, & x>0 \\[3pt] 0, & x<0 \\ \end{cases}$$

Spoiler

I'm aware that we can also redefine $|x|$ as $\sqrt{x^2}$, but I've seen this method used to differentiate $|x|$

Why does this work or not work? Is it because the piecewise function has its domain separated by inequalities ($x<0$ or $x>0$) whereas if there was an equality ($x=0$), this method fails?

 

[Euge]

Hi again Rido12,

The major issue in your argument is the statement "The question gives us $f(0) = 0$ when $x = 0$, so it follows that $f'(0) = 0$ when $x = 0$ because the derivative of $0$ is $0$". What does the derivative of $0$ being $0$ have to do with the value of $f'(0)$?

There's something else I need to add. Even before calculating $f'(0)$ you must check that $f$ is continuous at $0$. I assumed you already did this, but in case you haven't, use L'hospital's rule and continuity of $g'$ with $g'(0) = 0$ to get

$\displaystyle \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{g(x)}{x} = \lim_{x \to 0} g'(x) = g'(0) = 0 = f(0)$.

Thus, f is continuous at $0$.

 

[I like Serena]

That's true, but I'm getting tangled up here.

Isn't the way I'm getting $f'(x)$ analogous to what you would do with an absolute value function?

Not quite.
The absolute value function is not differentiable at 0, while it turns out that your function is.

$$f(x)=|x|=\begin{cases}x, & x>0 \\[3pt] -x, & x<0 \\ \end{cases}$$

Differentiating both parts of the piecewise function:

$$f'(x)=\begin{cases}1, & x>0 \\[3pt] -1, & x<0 \\ \end{cases}$$

Differentiating yet again:

$$f''(x)=\begin{cases}0, & x>0 \\[3pt] 0, & x<0 \\ \end{cases}$$

Spoiler

I'm aware that we can also redefine $|x|$ as $\sqrt{x^2}$, but I've seen this method used to differentiate $|x|$

Why does this work or not work? Is it because the piecewise function has its domain separated by inequalities ($x<0$ or $x>0$) whereas if there was an equality ($x=0$), this method fails?

All true.
However, in this case both $f'(0)$ and $f''(0)$ are undefined.

There's something else I need to add. Even before calculating $f'(0)$ you must check that $f$ is continuous at $0$.

Indeed, I wondered about this at first.
But doesn't the fact that $f'(0)$ turns out to exist, imply that $f$ is continuous at $0$?

 

[Dethrone]

Going back to the original, I read the piecewise function as the following:

When $x$ is any value but $0$, the function becomes $\frac {g(x)}{x}$. When $x$ is $0$, the function is $f(x)=0$.

The function $f(x)=0$, it is a constant function, where $f'(x)=0$. That is where I'm coming from. This is true for a general function where $f(x)=0$ is everywhere or for all $x$, but not true for a piecewise function where that is true only when $x=0$.

 

[I like Serena]

Going back to the original, I read the piecewise function as the following:

When $x$ is any value but $0$, the function becomes $\frac {g(x)}{x}$. When $x$ is $0$, the function is $f(x)=0$.

The function $f(x)=0$, it is a constant function, where $f'(x)=0$. That is where I'm coming from. This is true for a general function where $f(x)=0$ is everywhere or for all $x$, but not true for a piecewise function where that is true only when $x=0$.

If a function is only $0$ at $x=0$, it is not a constant function.

Suppose we define:
$$h(x) = \begin{cases} x & \text{if } x \ne 0 \\ 0 & \text{if } x = 0 \end{cases}$$
Then we still have that:
$$h'(0)=1 \ne 0$$

That is because $h'(0)$ is actually defined as:
$$h'(0) = \lim_{h\to 0} \frac{f(h) - f(0)}{h}$$
where $h\ne 0$. $h$ only approaches $0$. (Nerd)

 

[Dethrone]

Alright! It all makes sense now :D
Thanks everyone (Nod)

 

[Euge]

Indeed, I wondered about this at first.
But doesn't the fact that $f'(0)$ turns out to exist, imply that $f$ is continuous at $0$?

I meant to say "you should check" rather than "you must check". Sometimes I do that check when dealing with more complicated functions.