Is Rolle's theorem applicable to f(x)=tanx, when 0 < x < r??

1 Answer
Jun 8, 2015

No, since f(0)!=f(r) when 0 < r< pi/2. But the Mean Value Theorem is applicable when 0 < r < pi/2.

It says there is a number c (depending on r) in the interval (0,r) with the property that sec^{2}(c)=\frac{tan(r)-tan(0)}{r-0}=tan(r)/r (since sec^{2}(x)=d/dx(tan(x))), which is equivalent to rsec^{2}(c)=tan(r).

One thing this implies is that, since sec^{2}(c) = 1/(cos^{2}(c)) >1 when 0 < c < pi/2, it follows that tan(r)>r when 0 < r < pi/2. In other words, the graph of y=tan(x) is above the graph of y=x when 0 < x < pi/2.