Let \(f\) be the function that is differentiable on the open interval \((0,20)\). If \(f(5)=-10\), \(f(10)=10\), and \(f(15)=-10\), which of the following must be true?

• \(\hskip{1em}f\) has at least \(2\) zeroes.
• \(\hskip{1em}\)The graph of \(f\) has at least one horizontal tangent.
• \(\hskip{1em}\)For some \(c\), \((10\lt c \lt 15)\), \(f(c)=3\).

Summary Submit Skip Question

Sketch a curve through the three given points. From this, we can tell that all three statements are true.

You should try to disprove the statements by drawing different curves.

Statements I and III are guaranteed by the Intermediate Value Theorem.

Given \(f(5)=-10\) and \(f(10)=10\), there must be \(f(c)=0\).

Similarly, on the interval from point \(f(10)=-10\) and \(f(15)=10\), there must also be a zero.

Additionally, there must also be \(f(c)=3\) since \(3\) is between \(-10\) and \(10\).

We can use Mean Value Theorem or Rolle's Theorem to confirm statement II.

$$ f'(c) = \frac{f(b)-f(a)}{b-a} $$ $$ f'(c) = \frac{-10-(-10)}{15-5} $$ $$ f'(c) = \frac{0}{10} $$ $$ f'(c)=0 $$