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December 18, 2014

December 18, 2014

Posted by **Jenn** on Tuesday, March 26, 2013 at 10:53am.

f(x)=9x^4-3x^2+5x-1;[0,1]

- Use the intermediate value -
**bobpursley**, Tuesday, March 26, 2013 at 10:58amThere are a number of ways to do this, lets go to the idiot's guide to Math...

f(0)=-1

f(1)=9-3+5-1=10

so how can one get to 10 from -1 by not crossing the y=0 axis?

- Use the intermediate value -
**Jenn**, Tuesday, March 26, 2013 at 11:08amNot sure that is why I asked :)

- Use the intermediate value -
**Steve**, Tuesday, March 26, 2013 at 11:12amThe IVT is dependent on the fact that f(x) is continuous. That is, f(x)

**cannot**get from -1 to 10 without being 0 somewhere on the way.

If f is not continuous, then there might be a hole at f=0, so there would be no guarantee that f(c)=0 for some 0<c<1.

- Use the intermediate value -
**bobpursley**, Tuesday, March 26, 2013 at 11:12amReread the intermediate value theorem, it concludes that one can't get to 10 from -1 with a continuous function without passing the y=0 axis. Often, the mean value, and intermediate value theorem are written in math texts by lawyer want-to-be types, so complex, it loses its meaning.

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