`f(x) = sin^2(x) + sin(x)` Consider the function on the interval (0, 2pi).  Apply first derivative test to identify all relative extrema.

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Chapter 3, 3.3 - Problem 47 - Calculus of a Single Variable (10th Edition, Ron Larson).
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You need to evaluate the relative extrema of the function, hence, you need to find the solutions to the equation `f'(x) = 0` .

You need to determine the first derivative, using the chain rule, such that:

`f'(x) = (sin^2 x + sin x)' => f'(x) = 2sin x*cos x + cos x`

You need to solve for x the equation f'(x) = 0:

`2sin x*cos x + cos x = 0`

Factoring out cos x yields:

`cos x*(2sin x + 1) = 0`

`cos x = 0`

You need to remember that the cosine function is 0, in interval `(0,2pi), ` for `x = pi/2` and `x = 3pi/2` .

`2 sin x + 1 = 0 =>  2sin x = -1 => sin x = -1/2`

You need to remember that sine function is negative on intervals `(pi,3pi,2)` and `(3pi/2,2pi).`

`x = pi + pi/6 => x = (7pi)/6`

`x = 2pi - pi/6 => x = (11pi)/6`

Hence, evaluating the relative extrema of the function yields `((7pi)/6, f((7pi)/6))` and `((11pi)/6,f((11pi)/6)).`

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