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Question:

Let $f(x)=\begin{cases} 6, & x\le1\\7-x, & x\gt1\end{cases}$ then for $f(x), x=1$ is

A) a point of local maxima

B) a point of local minima

C) neither a point of local maxima nor minima

D) a stationary point

My Attempt:

The graph of $f(x)$ has a sharp edge at $x=1$. So, it's not a stationary point but a critical point.

Also, $f(1+h)\lt f(1)$, so, $x=1$ is not a point of minima.

And $f(1-h)$ is not less than $f(1)$, so, $x=1$ is not a maxima either. Or is it?

I am confused between A) and C).

If it is indeed C) then what will we call $x=1?$

If a critical point is neither maxima nor minima nor inflection then what is it?

aarbee
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1 Answers1

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Your function $f$ isn't differentiable at $x = 1$. A point $a$ in the domain of the function $f$ is a critical point iff $f'(a) = 0$ (Some people do define a critical point to be a point where either the derivative is zero or the function is not differentiable). In this case: $$\lim_{x \to 1^-} \frac{f(x)-f(1)}{x-1} = \lim_{x \to 1^-} 0 = 0$$ $$\lim_{x \to 1^+} \frac{f(x)-f(1)}{x-1} = \lim_{x \to 1^-} \frac{1-x}{x-1} = -1$$ Since these two derivatives don't coincide, the function isn't differentiable there. Now, observe that when $x > 1$, we have that $-x < -1$. So, $7-x < 6$. In other words, at $x = 1$, $f$ has a local maximum.

This will depend a little bit on your definition of local maximum. Let $f: I \to \mathbb{R}$ be a function defined on an open interval $I$ and let $a \in I$. Usually, we say that $a$ is a local maximum of $I$ iff there is a $\delta > 0$ such that $(a-\delta,a+\delta) \subseteq I$ and: $$\forall x \in (a-\delta,a+\delta): f(x) \leq f(a)$$ If you demand that the inequality be strict, then $x = 1$ is not a local maximum and certainly, it isn't a local minimum. You'll have to check your definitions to determine what's the answer in this case.

Mousedorff
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