True or False? #3

Calculus Level 4

There is a continuous function $f(x),$ defined in the set of all reals. For any real numbers $a$ and $b$ that satisfy $a<b,$ $f(x)$ always satisfies $f(a)>f(b).$

Which of the followings are correct?

A. $\displaystyle \lim_{h\to0} \frac{f(2+h)-f(2)}{h}\le0.$

B. There is always only one real root for $f(x)=0.$

C. There is always only one real root for $f(x)=f(-x+1).$

This problem is a part of <True or False?> series .

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

1 solution

Boi (보이)
Jul 25, 2017

A. (counterexample)

$f(x)=\cases{\begin{aligned} &-x &(x\ge2) \\ &-2x+2&(x<2) \end{aligned}}$

Then $\displaystyle \lim_{h\to0} \frac{f(2+h)-f(2)}{h}$ doesn't have a value, and therefore the statement is false.

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

B. (counterexample)

$f(x)=e^{-x}.$

Solution for $f(x)=0$ does not exist.

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

$f(x)$ monotonously decreases and $f(-x+1)$ monotonously increases.

And since it's clear that $y=f(x)$ and $y=f(-x+1)$ meets at point $\left(\dfrac{1}{2},~f\left(\dfrac{1}{2}\right)\right),$ they must meet at only one point.

Thus, the solution for $f(x)=f(-x+1)$ is only one, $x=\dfrac{1}{2}.$

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

From above, only C is correct.

True or False? #3

1 solution

0 pending reports