Consider the equation below. f(x) = 6 cos2x ? 12 sin x, 0 ? x ? 2? (a) Find the
ID: 2866548 • Letter: C
Question
Consider the equation below. f(x) = 6 cos2x ? 12 sin x, 0 ? x ? 2? (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) Correct: Your answer is correct. Find the interval on which f is decreasing. (Enter your answer in interval notation.) Correct: Your answer is correct. (b) Find the local minimum and maximum values of f. local minimum Correct: Your answer is correct. local maximum Correct: Your answer is correct. (c) Find the inflection points. (x, y) = (smaller x-value) (x, y) = (larger x-value) Find the interval on which f is concave up. (Enter your answer in interval notation.) Find the interval on which f is concave down. (Enter your answer in interval notation.)
Explanation / Answer
f(x) = 6cos(2x) - 12sin(x), 0 < x < 2pi
(a) Find the interval on which f is increasing. (Enter your answer in interval notation.)
So, we need the derivative
f'(x) = -12sin(2x) - 12cos(x) = 0
-12*2sinxcosx - 12cosx = 0
24sinxcosx + 12cosx = 0
12cosx(2sinx + 1) = 0
cosx = 0 , sinx = -1/2
x = pi/2 and 3pi/2 and x = 7pi/6 and 11pi/6 ---> Critical points
So, this splits number line into some regions, namely : (0 , pi/2) , (pi/2 , 7pi/6) , (7pi/6 , 3pi/2) , (3pi/2 , 11pi/6)
and (11pi/6 , 2pi)
And by checking, we get the answers :
Increasing : (pi/2 , 7pi/6) and (3pi/2 , 11pi/6) --> ANSWER
Decreasing : (0 , pi/2) , (7pi/6 , 3pi/2) and (11pi/6 , 2pi) ---> ANSWER
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b) Local max and min :
We have critical points, pi/2 , 7pi/6 , 3pi/2 and 11pi/6
We plug those into the original function
When x = pi/2 , y = -18 ---> local minimum
When x = 7pi/6 , y = 9 --> local maximum
When x = 3pi/2 , y = 6 --> local minimum
When x = 11pi/6 , y = 9 --> local maximum
So, local minimum values : -18 when x = pi/2 and 6 when x = 3pi/2
Local maximum values : 9 when x = 7pi/6 and 9 when x = 11pi/6
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(c) Find the inflection points.
f'(x) = -12sin(2x) - 12cos(x)
f''(x) = second derivative = -24cos(2x) + 12sin(x) = 0
-24(1 - 2sin^2(x)) + 12sinx =0
-2(1 - 2sin^2(x)) + sinx = 0
4sin^2(x) + sin(x) - 2 = 0
sin(x) = (-1 +/- sqrt(1 + 32)) / 8
sin(x) = (-1 + sqrt(33)) / 8 and (-1 - sqrt(33))/8
sin(x) = 0.5930703308172536 and sin(x) = -0.8430703308172536
So, x = 0.635 and 2.507 and x = 4.145 and 5.280
When x = 0.635, y = -5.340
When x = 2.507, y = -5.332
When x = 4.145, y = 7.586
When x = 5.280, y = 7.587
So, the above are the inflection points approximated to nearest thousandth.