Please show all work, and how you got the answer. Consider the equation: f(x) =

Question

Please show all work, and how you got the answer. Consider the equation: f(x) = 7 cos^2 x ? 14 sin x, 0 ? x ? 2? 1) Find the interval on which f is concave up. (Enter your answer in interval notation.) 2) Find the interval on which f is concave down. (Enter your answer in interval notation.)

Explanation / Answer

f'(x) = - 14 sin(x) cos(x) - 14 cos(x) = -14 cos(x) ( sin(x) + 1) So f'(x)=0 when : cos(x) = 0 => x = pi/2 & 3pi/2 & sin(x) = -1 => x = 3pi/2 when x x = 3pi/2 testing the sign of f''(x) around these points we get : there are two inflection points at x = pi/6 , 5pi/6 e) the curve is concave up in the interval (0 , pi/6 ) & ( 5pi/6 , 2pi) the curve is concave down in the interval (pi/6 , 5pi/6

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