Thread: Calculus Max and Min

Consider the function f(x)=sin(x)+(1/4)*sin(2x) on the interval [0,2pi].

Find numbers a, b and c so that a < b < c and f(x) is concave in one direction

on each of the intervals (0,a),(a,b),(b,c) and (c,2pi).

Originally Posted by Velvet Love

Consider the function f(x)=sin(x)+(1/4)*sin(2x) on the interval [0,2pi].

Find numbers a, b and c so that a < b < c and f(x) is concave in one direction

on each of the intervals (0,a),(a,b),(b,c) and (c,2pi).

$\displaystyle f'(x) = \cos{x} + \frac{1}{2}\cos(2x)$

$\displaystyle f''(x) = -\sin{x} - \sin(2x) = -\sin{x} - 2\sin{x}\cos{x} = 0$

$\displaystyle -\sin{x}(1 + 2\cos{x}) = 0$

set each factor equal to 0 and solve for x , then determine the concavity of each interval between the critical values.