Banach's fixed point theorem application

We want to show that the equation $f(x):={1\over 2}\cos ({\sin (x+2)+7x+1 \over 5}) =x$ has only one root. Since $f:R\to R$ and $R$ is a complete metric space, by Banach fixed point theorem we only need to show that $f$ is a contraction. We have $$ f'(x)={-1\over 2}{\cos (x+2)+7 \over 5}\sin ({\sin (x+2)+7x+1 \over 5}) \\ \; \\ \implies |f'|\le {1\over 2}{|\cos (x+2)|+7 \over 5}|\sin ({\sin (x+2)+7x+1 \over 5})|\le {1\over 2}{1+7 \over 5}={4\over 5}$$ So by mean value theorem $$ |f(x)-f(y)|=|f'(c)(x-y)|\le {4\over 5}|x-y|$$ which means $f$ is a contraction.