Prove x^3 = cosx + 16
Not sure how to go about this one? Any suggestions?
Having graphed it, I see that there is an intersection at x slightly larger than 2. Using that, I can give a more formal proof by noting that $\displaystyle 2^3= 8$ while $\displaystyle cos(2)+ 16$ is greater than 15.5 and that $\displaystyle 3^3= 27$ while $\displaystyle cos(3)+ 16$ is only slightly larger than 15.
That is, $\displaystyle 2^3< cos(2)+ 16$ while $\displaystyle 3^3> cos(3)+ 16$. Since $\displaystyle x^3$ and cos(x)+ 16 are continuous functions, there must be x between 2 and 3 such that $\displaystyle x^3= cos(x)+ 16$.