Thread: Prove using the mean value theorem

Using mean value theorem , prove that
for all

For all x in (0,pi/2) there is a c in (0,x) such that

[tan(x)-tan(0)]/[x-0] = sec^2(c) > 1 for c in (0,pi/2)

tan(x)/x > 1

tan(x) > x

Originally Posted by Calculus26

For all x in (0,pi/2) there is a c in (0,x) such that

[tan(x)-tan(0)]/[x-0] = sec^2(c) > 1 for c in (0,pi/2)

tan(x)/x > 1

tan(x) > x

I didnt get this part could u please explain

[tan(x)-tan(0)]/[x-0] = sec^2(c)

Originally Posted by zorro

I didnt get this part could u please explain

[tan(x)-tan(0)]/[x-0] = sec^2(c)

that is just applying the mean value theorem. [f(b)-f(a)] / (b-a) = f'(c), in this case a=0, b=x, and f(x) = tan(x) so f'(x) = sec^2(x).