Richard B. answered • 04/24/20
Electrical Engineering degree specializing in Physics, & Mathematics
Angie, first of all, understand that the problem is best solved in terms of radiance. Now let’s identify the first and second derivative right off the bat. First derivative is call it f’ = 7cox(x)-7sin(x) and the second derivative is f’’ = 7cos(x)-7sin(x). By setting the first derivative equal to zero and solving for X, we get 2 answers of course because the domain is between 0 and 2pi. These are called critical points at pi/4 and 5pi/4. So now we have to look at what happened to the left of pi/4 (a value around .8), between pi/4 and 5pi/4 (between about .8 and about 3.9), then to the right of 5pi/4. So here is what I get and you can plot these three functions (original function, first derivative, and second derivative) and see how these correspond to the conclusions below. So to the left of pi/4, the f' > 0, therefore the function is increasing. Between the two critical points the f'<0, therefore the function is decreasing. And to the right of 5pi/4, the f'>0 therefore the function is increasing again. We also note that the f'' = 0 at 3pi/4 (about 2.5) and 7pi/4 (5+). This gives us our inflection points so we can determine the function's concavity. So at the first point of 3pi/4 we notice that the f' <0, therefore the function is going from concave down to concave up. At the second point where the f'' = 0 7pi/4 (about 5+) is the second inflection point, at this point the f' > 0, therefore the function is going from concave up to concave down. And finally at the first critical point where the f’=0, and f’’< 0, therefore the function is at a max. Also note at f’=0, and f’’>0, therefore the function is at a minimum.
