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Question Number 122293 by mathocean1 last updated on 15/Nov/20

g i v e n f (x) = {c o s}^{2} x

f o r x \in [- \frac{π}{12}; \frac{π}{12}], ∣ f^{'} (x) ∣⩽ \frac{1}{2} .

1) s h o w t h a t \forall x, y \in [- \frac{π}{12}; \frac{π}{12}];

∣ {c o s}^{2} x - {c o s}^{2} y ∣⩽ \frac{1}{2} ∣ x - y ∣

Commented by ZiYangLee last updated on 16/Nov/20

By MVT, we have

\frac{f (x) - f (y)}{x - y} = f^{'} (c) ⩽ \frac{1}{2}

where c \in (x, y),

Hence,

∣ f (x) - f (y) ∣⩽ \frac{1}{2} ∣ x - y ∣

Answered by mathmax by abdo last updated on 15/Nov/20

\forall x \in [- \frac{π}{12}, \frac{π}{12}] ∣ f^{^{'}} (x) ∣⩽ \frac{1}{2} \Rightarrow - \frac{1}{2} ⩽ f^{^{'}} (x) ⩽ \frac{1}{2} \Rightarrow

- \frac{1}{2} \int_{x}^{y} dt ⩽ \int_{x}^{y} f^{^{'}} (t) dt ⩽ \frac{1}{2} \int_{x}^{y} dt \Rightarrow

- \frac{1}{2} (y - x) ⩽ f (y) - f (x) ⩽ \frac{1}{2} (y - x) \Rightarrow∣ f (x) - f (y) ∣⩽ \frac{1}{2} ∣ x - y ∣ \Rightarrow

∣ \cos^{2} x - \cos^{2} y ∣⩽ \frac{∣ x - y ∣}{2}