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I have discovered a series of calculus questions. However, I am not really equipped with the necessary knowledge to answer them. How should theorems like the IVP, mean value theorem, etc. be applied to answer these questions?

State whether the given situation is possible or impossible

Consider the interval to be $[a,b]$, $a>0$, and that $c$ is an element of interval $[a,b]$. Suppose also that each function $f$ is continuous over $[a,b]$:

  1. $f$ has a unique positive maximum, a unique negative minimum and two values $c$ such that $f(c)=0$

  2. $f$ has two maxima, two minima, and a unique value $c$ such that $f(c)=0$

  3. $f$ has exactly three values $c$ such that $f(c)=0$, its minimum is negative, its maximum is positive

  4. $f$ has exactly three values $c$ such that $f(c)=0$, its minimum is positive, its maximum is negative

Originally, the interval is $[a,a]$, which is most likely a typo.

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John Glenn
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  • Your intervals should not be $[a,a]$, which is only one point. – Ross Millikan Mar 14 '18 at 05:07
  • You must have meant $[a,b]$, not $[a,a]$. – symplectomorphic Mar 14 '18 at 05:08
  • Do you think it's a typo? This was how it was phrased. – John Glenn Mar 14 '18 at 05:08
  • They may have meant $[-a,a]$ with $a \gt 0$. The interval does not really matter as long as it is closed, connected, and larger than one point. – Ross Millikan Mar 14 '18 at 05:09
  • Okay thanks @RossMillikan. Do you think $f(c)=0$ should be $f'(c)=0$, too? – John Glenn Mar 14 '18 at 05:10
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    I would guess they meant $f(c)=0$. There are no derivatives elsewhere in the problem. We are not given that $f$ is differentiable. – Ross Millikan Mar 14 '18 at 05:12
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    Without differentiability, the IVT is your main tool. It is also important that on a closed interval the function attains its max and min. You need to think about whether you can make a function that meets the requirements. If you can, a sketch is often the easiest answer. If you claim it is impossible, you need to prove it. Case 4 is impossible-can you see why? One more difficult one to prove is here. We just saw this one again. – Ross Millikan Mar 14 '18 at 14:37

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You are supposed to think about how $f$ might meet the requirements. You could start with the interval and think about min/max at the endpoints. An example for 1 is below. It is just $x^3-x$ on $[-\frac 32,\frac 32]$. Yes there are three cases where the function is zero, but they did not say exactly two. You can change the function to make it exactly two if you want. How would you do so? enter image description here

Ross Millikan
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  • So I have to test, by trial and error, whether $f$ can exist? Suppose I do not find any $f$ which satisfies the conditions, how can I conclusively rule out the possibility that there still might be? – John Glenn Mar 14 '18 at 05:42
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    @JohnGlenn: when you are not able to find a possibility by graphing then you need to prove that the situation leads to a contradiction. For this you need to use the theorems like IVT. – Paramanand Singh Mar 14 '18 at 07:05