Question

calculus - mean value theorem
pic attached

Answer 1

Sqrt(cos(sin(x_1))-cos(sin(x_2)))^2 ≤ Sqrt(x_1-x_2)^2

Answer 2

nah i think you have to use mean value theorem here... so let f(x)=cos(sin(x)) hence from MVT, we get

Answer 3

do you have an idea @John_ES

Answer 4

Sorry, I am stumped haha.

Answer 5

Interesting problem, I was also thinking how to apply the mean value theorem to this problem, but I don't have a clear path. :)

Answer 6

well i let \[f(x)=\cos(\sin(x))\]

Answer 7

\[f'(x) = -\sin(\sin(x))\cos(x)\]

Answer 8

so from mean value theorem,
\[\frac{\cos(\sin( x_1)) -\cos(\sin( x_2))}{x_1-x_2}=-\sin(\sin(x_0))\cos(x_0)\]

Answer 9

would that help @John_ES

Answer 10

Yes, well, I was talking about what would be the following steps :).

Answer 11

Sure, doesn't apply here, but may be could you apply that sin and cosine functions are bounded by 1?

Answer 12

May be, using,
\[|\cos(x)|\leq1\\
|\sin(x)|\leq1\]
\[\left|{\cos(\sin(x_1))-\cos(\sin(x_2))}\right|=|\sin(\sin(x_0))\cos(x_0)({x_1-x_2})|\leq|x_1-x_2|\]
provided,
\[|\sin(\sin(x_0))\cos(x_0)|\leq1\]