I have two math problems that I've already done involving concavity and points of inflection and a plethora of other things, but some of my answers seem odd to me and I'd like to check them.
The first function is f(x) = x + cosx on the interval of [0, 2pi].
I found the first derivative, f'(x) = 1 - sinx, and after finding the critical number, x = (pi/2), and plugging values into f'(x) that are greater and less than the critical number, I found that the function has no relative extrema because it's continually increasing.
Next, I found f"(x) = -cosx. The critical numbers here were pi/2 and (3pi)/2, and after plugging various values into f"(x), one less than pi/2, one between the two critical numbers, and one greater than (3pi)/2, I found that the function is concave up from the interval (pi/2, (3pi)/2), and concave down from the intervals (negative infinite, pi/2) and ((3pi)/2, infinite). I also found two inflection points, (pi/2, pi/2), and ((3pi)/2, (3pi)/2) after plugging the two critical numbers of the f"(x) function into f(x).
I think I might have made a mistake somewhere, because I don't understand how ((3pi)/2, (3pi)/2) is an inflection point, because it doesn't look like the concavity changes there. Can someone verify this?
Alright, now there's one more function: f(x) = (x^2-4)^2. This is what I got:
f'(x) = 4x^3 - 16x. critical numbers are x = 0, x = 2, x = -2. Function increases from (-2, 0) and (2, infinite), decreases from (negative infinite, -2) and (0,2).
Next, I found f"(x) = 12x^2-16. This is where I'm second guessing myself, because I then found two critical numbers to be x = -1.154701 and x = 1.154701, which gives me inflection points of (-1.155, 7.111) and (1.155, 7.111), after rounding. That just doesn't seem right to me at all.
Help?
The first function is f(x) = x + cosx on the interval of [0, 2pi].
I found the first derivative, f'(x) = 1 - sinx, and after finding the critical number, x = (pi/2), and plugging values into f'(x) that are greater and less than the critical number, I found that the function has no relative extrema because it's continually increasing.
Next, I found f"(x) = -cosx. The critical numbers here were pi/2 and (3pi)/2, and after plugging various values into f"(x), one less than pi/2, one between the two critical numbers, and one greater than (3pi)/2, I found that the function is concave up from the interval (pi/2, (3pi)/2), and concave down from the intervals (negative infinite, pi/2) and ((3pi)/2, infinite). I also found two inflection points, (pi/2, pi/2), and ((3pi)/2, (3pi)/2) after plugging the two critical numbers of the f"(x) function into f(x).
I think I might have made a mistake somewhere, because I don't understand how ((3pi)/2, (3pi)/2) is an inflection point, because it doesn't look like the concavity changes there. Can someone verify this?
Alright, now there's one more function: f(x) = (x^2-4)^2. This is what I got:
f'(x) = 4x^3 - 16x. critical numbers are x = 0, x = 2, x = -2. Function increases from (-2, 0) and (2, infinite), decreases from (negative infinite, -2) and (0,2).
Next, I found f"(x) = 12x^2-16. This is where I'm second guessing myself, because I then found two critical numbers to be x = -1.154701 and x = 1.154701, which gives me inflection points of (-1.155, 7.111) and (1.155, 7.111), after rounding. That just doesn't seem right to me at all.
Help?
