# Show that a function has both positive and negative on the neighborhood of 0

Show the function: $$f(x)=4x\sin\left(\frac{1}{x}\right)-2\cos\left(\frac{1}{x}\right)+1 ,\text{ if }x\neq0$$ takes both positive and negative values on any open interval around $x=0$, i.e. on interval (-c,c).

There's hint that any such interval contains zeroes of both cos(1/x) and sin(1/x), but I'm not exactly sure how to come to this result. Thanks.

If $1/x$ is a multiple of $2\pi$, then $\sin(1/x)=0$ and $\cos(1/x)=1$, so $f(x)=-1$.
If $1/x$ is an odd multiple of $\pi$, then $\sin(1/x)=0$ and $\cos(1/x)=-1$, so $f(x)=3$.