Question

determine the absolute maximum and minimum values for f(x) = (sinx)/(2 + cosx) on the interval [0,2pi]

Answer 1

getting hung up on the algebra/trig solving

Answer 2

\(\large\color{black}{ \displaystyle f(x)=\frac{ \sin x }{2 + \cos x}}\) over the interval of \(\large\color{black}{ \displaystyle [0,2\pi]}\)

Answer 3

Find \(\large\color{black}{ \displaystyle f'(x)}\).
Set \(\large\color{black}{ \displaystyle f'(x)=0}\).

the x solutions will be your critical numbers.

Also critical numbers can be any numbers that are:
1. boundaries of a closed interval (like in your case it is \(\large\color{black}{ \displaystyle 0}\) and \(\large\color{black}{ \displaystyle2\pi}\))
2. x values at which \(\large\color{black}{ \displaystyle f'(x)}\), provided they are a part of \(\large\color{black}{ \displaystyle f(x)}\)'s domain.

if your critical number exceeds the interval , exclude it.

Answer 4

got the derivative of (2cosx + 1) / (2 + cosx)^2

Answer 5

for the zeros, I got 4pi/3 and 2pi/3 , but for where the graph is undefined, I'm having troubles

Answer 6

I set 0 = (2 + cosx)^2 , and am coming up with a sub-problem of -2 = cosx , do I need to continue?

Answer 7

in other words, looks like the denominator can never equal zero

Answer 8

wolfram gives me different answers than I come up with, I get +/- sqrt(3)/3 for the max/min, but wolfram is different, why http://www.wolframalpha.com/input/?i=absolute+max+and+min+of+f%28x%29+%3D+sinx+%2F+%282+%2B+cosx%29+between+0+and+2pi

Answer 9

so, for 2 + cosx there is no x val;ues that makes it zero.

Answer 10

the least for cos(x) is -1.

Answer 11

\(\large\color{slate}{ 2 + \cos x\ge1 }\)

Answer 12

you are right