Think Bolzano

Calculus Level 2

Fill in the blank.

Let $f(x)$ be a continuous real-valued function with $f(1) = -1$ and $f(4) = 3.$ If $1 < c < 4,$ then each of the following except for $\text{\_\_\_\_\_\_}$ is guaranteed to be a value of $f(c).$

0 1 2 4

2 solutions

Denton Young
Jun 16, 2017

By the intermediate value theorem, $f(x)$ takes all values between -1 and 3 on the interval [1, 4]. The only choice not in this range is 4.

It was a normal question. I was asked a question based on this concept in an interview.

Amit Kumar - 3 years, 11 months ago

Why couldn't the function reach a relative maximum of 4 somewhere between x =1 and x = 4 and then dip back down to 3 at x = 4?

Stephen Naus - 3 years, 11 months ago

It can. The way I originally wrote the question, it stated "Which of these values is NOT guaranteed to occur as x ranges from 1 to 4"? The staff edited the question to make it less correct, for some reason.

Denton Young - 3 years, 11 months ago

Huh that's strange. Thanks for the clarification though. Threw me for a loop for a second

Stephen Naus - 3 years, 11 months ago

I have re-edited the problem to make it correct again.

Denton Young - 3 years, 11 months ago

Thanks. A moderator has wrongly edited this problem. I see that the error has been rectified.

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the “line line line” menu in the bottom right corner. This will notify the problem creator (and eventually staff) who can fix the issues.

Brilliant Mathematics Staff - 3 years, 11 months ago

Just for completeness, a example of a function that does not take the value $4$ anywhere in the interval $[1,4]$ is $f(x)= \frac{4x-7}{3}$ .

Shourya Pandey - 3 years, 11 months ago

Function is wrong.....interchange 3 by 4

Nihal Patel - 3 years, 10 months ago

Emmanuel David
Oct 30, 2017

Think Bolzano

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