What are the critical values, if any, of # f(x)= x cos x −3 sin x +2x in [0,2pi]#?

Question

Emma Aldridge · Accepted Answer

#0#, #2pi# and a value near #4# Please see below.

#f'(x) = xsinx-2cosx+2# By inspection #x=0# and #x= 2pi# are solutions. #f'(pi) > 0# and #f'((3pi)/2) < 0# so there is another critical number in #[pi,(3pi)/2]#. Numerical or technological techniques find the solution at about #4#.

Ezra Adams · Answer

undefined To find the critical values of $ f(x) = x \cos x - 3 \sin x + 2x $ in the interval $[0, 2\pi]$, we first find its derivative and then solve for where the derivative equals zero or is undefined.

Taking the derivative of $ f(x) $ with respect to $ x $:

$ f'(x) = \frac{d}{dx}[x \cos x] - \frac{d}{dx}[3 \sin x] + \frac{d}{dx}[2x] $

$ f'(x) = \cos x - x \sin x - 3 \cos x + 2 $

Setting $ f'(x) $ equal to zero and solving for $ x $:

$ \cos x - x \sin x - 3 \cos x + 2 = 0 $

Combining like terms:

$ -2\cos x - x\sin x + 2 = 0 $

Now, there's no algebraic way to solve this equation for $ x $, so numerical methods or graphical methods are typically used to find approximate solutions within the interval $[0, 2\pi]$.

Once you've obtained these approximate solutions, you can check each one to see if it corresponds to a maximum, minimum, or inflection point by using the second derivative test or by examining the behavior of $ f'(x) $ around those points.