OK, thanks for the help with finding the actual derivative. So I set it = 0, which gave me
6cosx 4 = 0
After solving that, I was left with
cosx = -2/3
Now, would I have to take out the cosine by using the arc-cosine function, leaving me with x = .841? I plugged f'(x) into my calculator and it doesn't have a constant positive slope, so I definitely did something wrong. Please help!
2007-06-13 10:25:42 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It won't. You found x to be .841. That is the location where the derivative is 0. Your function has a horizontal tangent at that location.
If you want to check your point, plug in values on either side of .841 and make sure that the derivative is + on the left and - on the right (if you want a max)
and vice versa if you want a min.
2007-06-13 10:30:04 · answer #1 · answered by Anonymous · 0⤊ 1⤋
x = arccos -2/3 = 131.81 degrees = 2.3 radians
x also = 228.19 degrees = 3.98 radians sice the cosine is negative in quadrant II and III.
2007-06-13 10:40:23 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋