Math  /  Calculus

QuestionConsider the equation below. (If an answer does not exist, enter DNE.) f(x)=6cos2(x)12sin(x),0x2πf(x)=6 \cos ^{2}(x)-12 \sin (x), \quad 0 \leq x \leq 2 \pi (a) Find the interval on which ff is increasing. (Enter your answer using interval notation.) (π2,3π2)\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)
Find the interval on which ff is decreasing. (Enter your answer using interval notation.) (0,π2)(3π2,2π)\left(0, \frac{\pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right) (b) Find the local minimum and maximum values of ff. local minimum value 12\quad-12 local maximum value 12 (c) Find the inflection points. (Order your answers from smallest to largest xx, then from smallest to largest yy.) (x,y)=(π6,32)(x,y)=(5π6,32)\begin{array}{l} (x, y)=\left(\frac{\pi}{6}, \frac{3}{2}\right) \\ (x, y)=\left(\frac{5 \pi}{6}, \frac{3}{2}\right) \end{array}
Find the interval on which ff is concave up. (Enter your answer using interval notation.) (0,π6)(5π6,3π2)\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, \frac{3 \pi}{2}\right)
Find the interval on which ff is concave down. (Enter your answer using interval notation.)

Studdy Solution
Identify inflection points where f(x)=0 f''(x) = 0 and changes sign.
Inflection points occur at:
x=π6,5π6 x = \frac{\pi}{6}, \frac{5\pi}{6}
Evaluate f(x) f(x) at these points:
f(π6)=6cos2(π6)12sin(π6)=32 f\left(\frac{\pi}{6}\right) = 6 \cos^2\left(\frac{\pi}{6}\right) - 12 \sin\left(\frac{\pi}{6}\right) = \frac{3}{2}
f(5π6)=6cos2(5π6)12sin(5π6)=32 f\left(\frac{5\pi}{6}\right) = 6 \cos^2\left(\frac{5\pi}{6}\right) - 12 \sin\left(\frac{5\pi}{6}\right) = \frac{3}{2}
Inflection points: (π6,32)(\frac{\pi}{6}, \frac{3}{2}) and (5π6,32)(\frac{5\pi}{6}, \frac{3}{2})
The interval on which f f is concave down is (π6,5π6)(3π2,2π) \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right) .

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