f(x) = 9 sin x + 9 cos x, 0 ? x ? 2? (a) Find the interval on which f is increas

Question

f(x) = 9 sin x + 9 cos x, 0 ? x ? 2?
(a) Find the interval on which f is increasing. (Enter your answer using interval notation.)


Find the interval on which f is decreasing. (Enter your answer using interval notation.)


(b) Find the local minimum and maximum values of f.
local minimum value
local maximum value


(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 using interval notation.)


Find the interval on which f is concave down. (Enter your answer using interval notation.)

Explanation / Answer

Given, f(x) = 9sin(x) + 9cos(x) ; 0<= x <= 2

f'(x) = 9[cos(x) - sin(x)] and,

   f''(x) = -9[cos(x) + sin(x)]

(a)  For f(x) to be strictly increasing,

f'(x) > 0

i.e. cos(x) - sin(x) > 0

i.e. cos(x) > sin(x)

i.e. x [0, /4) U (5/4, 2]

For f(x) to be strictly decreasing,

f'(x) < 0

i.e. cos(x) - sin(x) < 0

i.e. cos(x) < sin(x)

i.e. x (/4, 5/4)

(b) For local maxima and minima, f'(x) = 0

i.e. cos(x) = sin(x)

i.e. tan(x) = 1

i.e. x = /4, 5/4

At x = /4, f''(x = /4) = -92 < 0    x = /4 is a local maxima and f(/4) = 92

At x = 5/4, f''(x = 5/4) = 92 > 0    x = 5/4 is a local minima and f(5/4) = -92

(c) For points of inflection, f''(x) = 0

i.e. cos(x) + sin(x) = 0

i.e. tan(x) = -1

i.e. x = 3/4, 7/4 and f(3/4) = 0 = f(7/4)

Points of inflection are (3/4 , 0) and (7/4 , 0)

(d) For the curve to be concave down, f''(x) < 0

i.e. -9[cos(x) + sin(x)] < 0

i.e. cos(x) + sin(x) > 0

   x (0 , /4) U (5/4 , 2)

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