Answer to Question #121876 in Calculus for desmond

Question

the function f : x ı→ a + b cos x, is defined for 0 ≤ x ≤ 2π. Given that f(0) = 10 and that f(2π/3) = 1 , find (i) the values of a and b, (ii) the range of f, (iii) the exact value of f(5π/6)

Answer 1

I. $f(x)=a+b cos x, 0\le x \le 2\pi.$

$f(0)=10,$ hence $a+b cos 0=10, a+b\cdot 1=10, a+b=10.$

$f(\frac{2\pi}{3})=1,$ hence $a+b cos \frac{2\pi}{3}=1, a+b\cdot (-\frac{1}{2})=1,$ when multiply this equation by 2, we have $2a-b=2.$

We have the system of linear equations $\begin{cases} a+b=10, \\ 2a-b=2. \end{cases}$

If we add the equations, we have $3a=12,$ hence $a=4.$

Then $4+b=10, b=6.$

Answer. $a=4, b=6.$

II. $f(x)=4+6 cos x.$

The magnitude of the function is 6.

The lower bound of the range for the function is found when $cos x=-1,$ and it's equal to $y=4+6\cdot (-1)=-2.$

The upper bound of the range for the function is found when $cosx=1$, and it's equal to $y=4+6\cdot 1=10.$

Hence the range of the f is $-2 \le y \le 10,$ interval notation $[-2, 10].$

Answer. $[-2, 10].$

III. $f(\frac{5\pi}{6})=4+6cos\frac{5\pi}{6}=4+6\cdot (-\frac{\sqrt{3}}{2})=4-3\sqrt{3}.$

Answer. $4-3\sqrt{3}.$