Show that this graph isn't differentiable at x=1

[opus]

Homework Statement

Sketch the graph of the following function and use the definition of the derivative to show that the function is not differentiable at x=1.

$$f(x) = \begin{cases} {-x^2+2} & \text{if } x \leq 1 \\ x & \text{if } x > 1 \end{cases}$$

Homework Equations

Derivative: $$f'(x) = \lim_{h \rightarrow 0} {\frac{f(x+h) - f(x)}{h}}$$

The Attempt at a Solution

Now I graphed this peicewise function and there is a sharp kink at x=1.
Visually, I know that there is no derivative here, because if we think about tangent line to the graph at the point of the kink, it can swivel around and there is no definite derivative.
Algebraically, I'm not sure what to say.

The point is continuous, but a point can be continuous and not be differentiable, but not the other way around.
I took the derivative function at $x\leq 1$ and that derivative function is $f'(x)= -2x$
I took the derivative function at $x>1$ and that derivative function is $f'(x) = 1$

So this tells me that the derivative functions are different at the left of x=1 and at the right, which is obvious because the piecewise has two separate function on these intervals anyways.

So I'm at a standstill on what I need to say mathematically to show that the function is not differentiable at x=1.

 

[fresh_42]

It is the same reason as before with your points of continuity: A limit (without specification from which side) exists, if it is equal from left and from right. This is not the case for the limit of the first derivative at $x=1$, therefore it is not differentiable here. A unique limit cannot be chosen, and thus it doesn't exist.

 

[opus]

This is not the case for the limit of the first derivative at x=1

Why doesn't a limit exist at x=1? It appears that from both sides, the graph approaches 1.

 

[opus]

See image

 

[fresh_42]

Why doesn't a limit exist at x=1? It appears that from both sides, the graph approaches 1.

Sure, but not the limit as it is in the definition of a derivative: $\lim_{h \to +0} \dfrac{f(1+h)-f(1)}{h} \neq \lim_{h \to -0} \dfrac{f(1+h)-f(1)}{h}\,.$

 

[Ray Vickson]

Homework Statement

Sketch the graph of the following function and use the definition of the derivative to show that the function is not differentiable at x=1.

$$f(x) = \begin{cases} {-x^2+2} & \text{if } x \leq 1 \\ x & \text{if } x > 1 \end{cases}$$

Homework Equations

Derivative: $$f'(x) = \lim_{h \rightarrow 0} {\frac{f(x+h) - f(x)}{h}}$$

The Attempt at a Solution

Now I graphed this peicewise function and there is a sharp kink at x=1.
Visually, I know that there is no derivative here, because if we think about tangent line to the graph at the point of the kink, it can swivel around and there is no definite derivative.
Algebraically, I'm not sure what to say.

The point is continuous, but a point can be continuous and not be differentiable, but not the other way around.
I took the derivative function at $x\leq 1$ and that derivative function is $f'(x)= -2x$
I took the derivative function at $x>1$ and that derivative function is $f'(x) = 1$

So this tells me that the derivative functions are different at the left of x=1 and at the right, which is obvious because the piecewise has two separate function on these intervals anyways.

So I'm at a standstill on what I need to say mathematically to show that the function is not differentiable at x=1.

You have already shown it in the previous paragraph, because you have shown that your limiting ratio does not have a limit as $h \to 0$: it matters whether you have $h \downarrow 0$ or $h \uparrow 0.$

 

[opus]

Let me see if I understand:

Using the derivative function, if x approaches 0 from the left, then x is less than 1. In this case, we use $f(x) = -x^2+2$
Taking the derivative of this, the slope (or derivative) as h approaches 0 from the left is -2

If x approaches zero from the right, the we are greater than 1. In this case, we use $f(x) = x$
Taking the derivative of this, the slope as h approaches 0 from the right is 1.

These individual derivates, from the left and right, are not equal, so the point at x=1 is indifferentiable.

 

[opus]

Shown work

 

[opus]

your limiting ratio does not have a limit as h→0

What do you mean by this? Does it have to do with just trying to take the derivative at x=1? Because I tried to do that, and couldn't manage to get one. I kept getting $\frac{0}{0}$ which is of course undefined.

 

[Merlin3189]

Just some very naive queries from a non-mathematician reading this.

Can this be true?

... $\lim_{h \to +0} \dfrac{f(1+h)-f(1)}{h} \neq \lim_{h \to -0} \dfrac{f(1+h)-f(1)}{h}\,.$

They look identical to me.
Presumably the f(x) on the left is different from the f(x) on the right. Is there some notation which makes this clear for "split" functions like f(x)?
Maybe $f_{x ≤ 1}(x) \ and f_{>1}(x)$
Or perhaps something like, $f(x) = \begin{cases} {g(x)=-x^2+2} & \text{if } x \leq 1 \\h(x)= x & \text{if } x > 1 \end{cases}$
so that you could write, say,
$\lim_{d \to +0} \dfrac{g(1+d)-g(1)}{d} \neq \lim_{d \to -0} \dfrac{h(1+d)-h(1)}{d}\,.$

From post #1
It seems to me that, for the simple polynomial type functions that I met at school, the derivative could equally be obtained using
$f'(x) = \lim_{h \rightarrow 0} {\frac{f(x) - f(x-h)}{h}}$
as well as the formula given in OP.
Is there a strict mathematical definition for derivatives that requires the stated formula?

 

[fresh_42]

Shown work

The second one looks o.k. as far as I can tell from the photo. The first one, however, looks wrong. For $x>1$ we have $\lim_{h \downarrow 0}\dfrac{f(1+h)-f(1)}{h}=\lim_{h \downarrow 0}\dfrac{1-1}{h}=0$ since the nominator is equally zero and the denominator only approaching it. There is no $\frac{0}{0}\,.$

 

[opus]

The second one looks o.k. as far as I can tell from the photo. The first one, however, looks wrong. For $x>1$ we have $\lim_{h \downarrow 0}\dfrac{f(1+h)-f(1)}{h}=\lim_{h \downarrow 0}\dfrac{1-1}{h}=0$ since the nominator is equally zero and the denominator only approaching it. There is no $\frac{0}{0}\,.$

For the second, as you had shown, I tried some voodoo by multiplying by the conjugate of the numerator to the top and bottom. In doing this, I got the limit as 1 and that made sense because to the right of one, it's the line y=x.
Without doing this however, I did get the limit to be 0, but that doesn't make sense at all by looking at the graph. Could you explain?

 

[fresh_42]

Presumably the f(x) on the left is different from the f(x) on the right. Is there some notation which makes this clear for "split" functions like f(x)?

I used the notation $h \to +0$ for the approach from the right and $h \to -0$ from the left. @Ray Vickson wrote $h \downarrow 0$ resp. $h \uparrow 0$ instead. Both is used and a matter of taste. Some also write $h \nwarrow 0$ or $h \searrow 0$ resp. $h \nearrow 0\,.$ There is no clear standard.

As we have only one function, I do not like to name the left and right branch differently. From a logical point of view, there is only $x \longmapsto f(x)$, no $g(x)$ and no $h(x)$.

 

[fresh_42]

For the second, as you had shown, I tried some voodoo by multiplying by the conjugate of the numerator to the top and bottom. In doing this, I got the limit as 1 and that made sense because to the right of one, it's the line y=x.
Without doing this however, I did get the limit to be 0, but that doesn't make sense at all by looking at the graph. Could you explain?

Sorry, my fault. I mistakenly took $f(x)=1$ for $x>1$ instead of $f(x)=x$. Here's the correction:

The second one looks o.k. as far as I can tell from the photo. The first one, however, looks wrong. For $x>1$ we have $\lim_{h \downarrow 0}\dfrac{f(1+h)-f(1)}{h}=\lim_{h \downarrow 0}\dfrac{1+h-1}{h}=\lim_{h \downarrow 0} \dfrac{h}{h}=1$ since the nominator is equal to the denominator. Both are simultaneously approaching $0$. There is no $\dfrac{0}{0}\,.$ as we never get there.

 

[fresh_42]

It seems to me that, for the simple polynomial type functions that I met at school, the derivative could equally be obtained using $f'(x) = \lim_{h \rightarrow 0} {\frac{f(x) - f(x-h)}{h}}$ as well as the formula given in OP.

That's the same thing. Maybe left and right will be different in the $h \to +0$ notation, as $h$ is now negative, but this doesn't matter - because they have to coincide from left and right!

 

[Merlin3189]

Thanks for the clarification. I'd noticed the up/down approaches in Ray's post, but missed it in yours. Blame my eyesight: perhaps the astigmatism likes vertical arrows but not horizontal ones.

 

[opus]

So does this make sense? Trying to visualize this.

In the top graph, the limit of the derivative from the left is -2 and from the right is 1. AT x=1 it gets wonky because there’s not a smooth transition.

In the top graph, which is a somewhat equivalent, there is a smooth transition so the different slopes on each side of x=1 converge on a derivative or slope of 0.

 

[fresh_42]

So does this make sense? Trying to visualize this.

In the top graph, the limit of the derivative from the left is -2 and from the right is 1. AT x=1 it gets wonky because there’s not a smooth transition.

In the top graph, which is a somewhat equivalent, there is a smooth transition so the different slopes on each side of x=1 converge on a derivative or slope of 0.

Yes, this makes sense. For short: $x \longmapsto |x|$ is not differentiable at $x=0\,.$ whereas $x \longmapsto x^n$ are.

 

[opus]

Ok great.

So to prove that the graph is not differentiable at x=1,

• Take the derivative from the left of x=1 as x gets very close to 0. This will give a rate of change, or derivative as x gets close to 1 from the left.
• Take the derivative from the right of x=1 as x gets very close to 0. This will give a rate of change, or derivative as x gets close to 1 from the right.
• These different intervals do not approach the same rate of change, or derivative, as they approach the same x value. This implies a kink in the graph at x=1, which coincides with what I drew.
  
• From this information we can say that f(x) is not differentiable at x=1

Would you change anything to this statement?

 

[fresh_42]

Ok great.

So to prove that the graph is not differentiable at x=1,

• Take the derivative from the left of x=1 as x gets very close to 0. This will give a rate of change, or derivative as x gets close to 1 from the left.
• Take the derivative from the right of x=1 as x gets very close to 0. This will give a rate of change, or derivative as x gets close to 1 from the right.
• These different intervals do not approach the same rate of change, or derivative, as they approach the same x value. This implies a kink in the graph at x=1, which coincides with what I drew.
  
• From this information we can say that f(x) is not differentiable at x=1

Would you change anything to this statement?

Yes. I would say: Calculate the limits (which would define a derivation) from the left and from the right as the derivative does not exist, the one-sided limits do! What you did in this statement, is to pretend you have two different differentiable functions with different derivatives at the point in question. This leads to the same result in this case, but is not what's going on. We have only one function and no derivative at $x=1$.

 

[opus]

Ok great, thank you everyone for your help.

 

[Ray Vickson]

Let me see if I understand:

Using the derivative function, if x approaches 0 from the left, then x is less than 1. In this case, we use $f(x) = -x^2+2$
Taking the derivative of this, the slope (or derivative) as h approaches 0 from the left is -2

If x approaches zero from the right, the we are greater than 1. In this case, we use $f(x) = x$
Taking the derivative of this, the slope as h approaches 0 from the right is 1.

These individual derivates, from the left and right, are not equal, so the point at x=1 is indifferentiable.

Exactly. So the ratio
$$\frac{f(1+h)-f(1)}{h}$$ will have two different limits: one for $h \to 0$ from above, and another for $h \to 0$ from below. That means that a limit as $h \to 0$ does not exist, because limits are not supposed to be different for the two different ways of having $h$ go to 0.

 

[opus]

Excellent, thank you Ray!