A calculus problem by Julian Poon

Calculus Level 3

A function $f(x)$ is described as such:

$f(x)$ is continuous for all non-zero real $x$
$f(x)$ is strictly increasing when $x>0$
$f(x)$ is strictly decreasing when $x<0$

Which of the following options must be true?

None of the others

f(x)

achieves its minimum at

x=0

f(x)

achieves its maximum at

x=0

\displaystyle f(x)>\lim_{ x\rightarrow 0 }{ f(x) }

\displaystyle f(x)<\lim_{ x\rightarrow 0 }{ f(x) }

1 solution

Richard Costen
Jun 27, 2017

Since the function definition does not specify anything about the function at $x=0$ , it can have any value or not even be defined at $x=0$ , thus it does not necessarily achieve a maximum or a minimum at this point. That eliminates 2 answers. Also, the left-hand limit does not need to equal the right-hand limit - there can easily be a jump at $x=0$ , thus by definition, $\displaystyle \lim_{x\rightarrow 0}{f(x)}$ may not even exist. The other 2 answers are thus eliminated.

The conditions state that f is continuous on non negative reals and this includes at 0, so both f(0) and the limit at 0 must exist and be equal to each other. Either your solution must be amended by saying that f doesn't need to be defined in the negative neighborhood around 0 allowing for the four statements to be untrue, or the condition in the problem must be tweaked to have continuity only on the positive reals.

Anthony Holm - 3 years, 11 months ago

I edited my solution a few days ago after the original problem was modified. Does it look correct now?

Richard Costen - 3 years, 11 months ago

A calculus problem by Julian Poon

1 solution

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