Christmas Streak 37/88: Think, Think, Think...

Calculus Level 5

For a function $f(x):\mathbb{R}\to\mathbb{R},$ which of the followings are true?

A. Even if $f$ is a bijective function, $f(x)$ can be discontinuous.

B. If some reals $p$ and $q$ satisfy $p< q$ and $f(p)f(q)<0,$ there exists a real number $x$ such that $p<x<q$ and $f(x)=0.$

C. If $f(2)=f(5),$ there exists some $x$ such that $2<x<5$ and $f'(x)=0.$

Notation: $\mathbb{R}$ denotes the set of reals.

This problem is a part of <Christmas Streak 2017> series .

Only A Only A and B Only B None of them Only C Only B and C Only C and A All of them

1 solution

Boi (보이)
Nov 4, 2017

$f(x)=\cases{\begin{aligned}& \dfrac{1}{x}~&(x \neq 0)\\ \\ & 0&(x=0)\end{aligned}}$ satisfies the condition.

$\therefore~\boxed{\text{TRUE}}$

B. (counterexample)

Consider a function $f(x)=\cases{\begin{aligned} & -2 ~& (x<0)\\ & \\ & 2 & (x\ge0)\end{aligned}}.$

$f(-1)f(1)<0$ but there is no such $x$ that $f(x)=0.$

$\therefore~\boxed{\text{FALSE}}$

C. (counterexample)

Consider a function $f(x)=\dfrac{1}{(2x-7)^2}.$

$f(2)=f(5),$ but there is no such $x$ that $f'(x)=0.$

$\therefore~\boxed{\text{FALSE}}$

From above, only A is correct.

@H.M. 유 your counterexample in part B does not satisfy the following condition of the function:

If $a<b$ then $f(a) > f(b)$

The above condition states that the function is a decreasing one . However , $x+2$ is an increasing function.

Moreover if the given condition holds how can $f(2) = f(5)$ .

May be you wanted to write the following : If $a<b$ then $f(a) \ge f(b)$

If I am wrong anywhere please tell me.

A Former Brilliant Member - 3 years, 7 months ago

I've totally editted the question. Thanks for pointing out.

Boi (보이) - 3 years, 7 months ago

@H.M. 유 Can you please reword statement A? It's very confusing; "even if there's only one value for $x$ for the equation $f(x)=t$ for any real number $t$ " seems to say that $x$ cannot have more than one value (which seems obvious since $f$ is a function) but also that $x$ is everywhere defined to be $t$ , a constant real number. I did not know that you could define $f(x)=\frac{1}{x}$ instead of $f(x)=t$

Alexander Ilmexia - 3 years, 4 months ago

I edited the "any" bit to "every". Does that make it a bit clearer?

Boi (보이) - 3 years, 4 months ago

I think you should change it to: "Even if $f(x)$ has values defined for all real numbers $x$ , $f(x)$ can still be discontinuous."

Alexander Ilmexia - 3 years, 4 months ago

@Alexander Ilmexia – I think it can be a bit more clearer by just stating that f is bijective.

Boi (보이) - 3 years, 4 months ago

Christmas Streak 37/88: Think, Think, Think...

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