Use the Intermediate Value Theorem to show that there is a cylinder of height h and radius less than r whose volume is equal to that of a cone of height h and radius r.(adsbygoogle = window.adsbygoogle || []).push({});

IVT states that: if f is continuous on a closed interval [a,b] and k is any number between f(a) and f(b), inclusive, then there is at least one number x in the interval [a,b] such that f(x)=k.

The volume of the cylinder is [tex] \pi x^2h[/tex] and the volume of the cone is [tex]\frac{\pi}{3} r^2h[/tex], where x< r. If r > x then, the curve [tex] \frac{\pi}{3} r^2h[/tex] is streched more than the curve [tex] \pi x^2h[/tex], i.e it is closer to the y-axis than the curve [\pi x^2h[/tex], so I cannot find k. And is not a<0 and b>r? Then f(a)<0 and f(b)>f(r)?I am studing this on my own. Please help. Thanks.

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# Show using the intermediate value theorem

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