How can a point of symmetry have a slope which isn't either $0$ or $±\infty$?

Question

I've got a bit of a doubt with a question I'm solving. It goes like this:

For $a>0$, let $f:[-4a,4a] \to R$ be an even function such that $f(x) = f(4a - x) \hspace{2 mm}\forall x \in [2a, 4a]$ and $$\lim_{h\to0} \frac{f(2a + h) - f(2a)}{h} = 4,$$ then $$\lim_{h\to0} \frac{f(h-2a) - f(-2a)}{2h} = \hspace{2mm}?,$$

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Here's my reasoning:

It's been given that the function is even, so it's obviously symmetric about the y-axis.
Another piece of information which has been give is that $f(x) = f(4a - x) \hspace{2 mm}\forall x\in [2a, 4a]$. So it's gotta be symmetric about the line $x = 2a$ in the interval $ x\in[0,4a]$ too.

So the graph would look something like this:

Either that, or a cusp or a sharp corner would exist at $x = ±2a$.

But, in the question, it's given that $$\lim_{h\to0} \frac{f(2a + h) - f(2a)}{h} = 4,$$

which means that the slope of the graph at $x = 2a$ is $4$.

How is this possible, if the only options are:

• as in the graph, a maxima (or a minima) exists, thus the slope is $0$.
• a sharp corner exists - which means that the derivative isn't defined at the point.
• a cusp exists - which means the slope is either $±\infty$!

So how is it possible that a portion of the graph be symmetric about a value, but still have a derivative which isn't either $0$ or $±\infty$? Or is it that the question is wrong itself?

Answer 1

Okay, thanks to Daniel Fischer I now know what's the problem with the question. They should have specified that $h>0$. With that cleared, the graph of the function would have been something like:

The right hand derivative of $x=2a$ is 4, which implies that the left side derivative will be $-4$ (because of the symmetry). Again, symmetry implies the right hand derivative of $x = 2a$ will be the negative of the left hand derivative $x = -2a$, which is -(-4) = 4.

The answer = $$\frac{rhd}{2} = 2.$$

Thanks, guys!

Answer 2

Since $f(4a-x)=f(x)$, letting $x=2a+h$ gives $f(2a+h)=f(2a-h)$. Now, since the function is even, we have $f(-2a)=f(2a)$. Then we get $$\frac{{f\left( {2a + h} \right) - f\left( {2a} \right)}}{h} = \frac{{f\left( {2a - h} \right) - f\left( {2a} \right)}}{h}$$ so $$\begin{align} \lim\limits_{h \to 0} \frac{{f\left( {2a + h} \right) - f\left( {2a} \right)}}{h} &= - \mathop \lim\limits_{h \to 0} \frac{{f\left( {2a - h} \right) - f\left( {2a} \right)}}{{ - h}} \\ f'\left( {2a} \right) &= - f'\left( {2a} \right) \\ f'\left( {2a} \right) &= 0 \end{align}$$

Thus if the derivative at $2a$ exists, we must have $f'(2a)=0$. This shows that $f'(2a)=4$ is impossible.

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The "centered" derivative relies on the fact that if the derivative exists $$\displaylines{ \mathop {\lim }\limits_{h \to 0} \frac{{f(a + h) - f\left( a \right)}}{h} = f'\left( a \right) \cr \mathop {\lim }\limits_{h \to 0} \frac{{f\left( a \right) - f(a - h)}}{h} = f'\left( a \right) \cr} $$ But summing gives $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f(a - h)}}{{2h}} = f'\left( a \right)$$

Answer 3

Good thinking. If you use the centered expression of the derivative $$f'(2a)=\lim_{h \to 0}\frac{f(2a+h)-f(2a-h)}{2h}$$ and the fact that $4a-(2a-h)=2a+h$ the numerator is zero.

Answer 4

There is no such function. For $0< h\leq 2a$ one has $${f(2a+h)-f(2a)\over h}={f(2a-h)-f(2a)\over h}=-\ {f\bigl(2a+(-h)\bigr)-f(2a)\over(-h)}\ .$$ Letting $h\to{0+}\>$ implies $4=-4$.