Find the range of $f:x \mapsto a+b\cos x$ The function $f:x \mapsto a+b\cos x$, is defined for $0 \le x\le 2\pi$. Given that $f(0) = 10$ and $f\left(\frac{2}{3}\pi\right) = 1$, find the values of $a$ and $b$, the range of $f$, and the exact value of $f\left(\frac{5}{6}\pi\right)$.
I was able to get the value of a and b given that $f(0) = 10$ and $f\left(\frac{2}{3}\pi\right) = 1$ which gives $a + b\cos (0) = 10$ and $a + b\cos\left(\frac{2}{3}\pi\right) = 1$. Therefore $a = 4$ and $b = 6$.
For the exact value of $f\left(\frac{5}{6}\pi\right)$; Since $a = 4$ and $b = 6$, $f(x) = a + b\cos x \Rightarrow  4 + 6\cos x$, Therefore $f\left(\frac{5}{6}\pi\right) = 4 + 6\cos\left(\frac{5}{6}\pi\right) = 4 + 6(-\frac{\sqrt3}{2})$ = $4-3\sqrt{3}$ or $-1.1962$.
Getting the range of f is what I don't know how to go about, can anyone here offer an explanation or answer? Thanks

Your calculation for $a,b$ is correct, and so is the calculation of $f\left(\frac56\pi\right)$.
For calculating the range of $f$, consider the following hint:

*

*For all $x$, you $-1\leq \cos x \leq 1$.

*The bounds above are strict, i.e., there exist values of $x$ for which $\cos x$ reaches values of $-1$ and $1$.

