Question

At how many points on the interval -2pi to 2pi does the tangent to the graph of the curve y=xcosx have slope pi/2

i got the derivative to xcosx as xsinx+cosx... now do i set that equal to pi/2?
if so how do you solve that because i dunno

Answer 1

the derivative is cosx - xsinx

Answer 2

thanks you! but what do i do after that?

Answer 3

first note that f(x)=f(-x), for f(x)=cosx - xsinx
consider only the positive x values, and multiply the final answer by 2
there is guaranteed to be at least one positive x solution, since at f(3pi/2)=3pi/2 > pi/2
and f(0)=1
and the function is continuous

Answer 4

*for the first part, it should say f(pi)=-1 instead of f(0)=1

Answer 5

the actual values of x are not easily solvable; the number of values is

Answer 6

in summary
f(0)=1
f(pi/2)=-pi/2
f(pi)=-1
f(3pi/2)=3pi/2
f(2pi)=1
the only domains we care about are (pi,3pi/2) and (3pi/2,2pi). f in these domain will hit pi/2
to verify that f only hits pi/2 once in each domain, we can compute
f'(0)=0
f'(pi/2)=-2
f'(pi)=pi
f'(3pi/2)=2
f'(5pi/3)=sqrt(3)-5pi/6
f'(2pi)=-2pi
from pi to 3pi/2, f' is positive thus f is increasing
from 5pi/3 to 2pi, f' is negative thus f is decreasing

thus the number of points is 4