Let f be the function defined by f(x) = sin^2x - sinx for x is between 0 and 3pi/2
a. find the x intercepts of f
b. write the equation of any horizontal tangents of f
c. find the intervals on which f is incereasing
d. sketch an accurate graph of f'(x) and explain how that graph can help you answer (c).
not sure if you've gotten the rest of it yet:
b) horizontal tangents exist where the derivative of the equation is equal to zero.
f(x) = sin(x)^2 - sin(x)
f'(x) = sin(2x) - cos(x)
solve for f'(x) = 0
sin(2x) - cos(x) = 0
sin(2x) = cos(x): since sin (2x) = 2sin(x)cos(x),
2sin(x)cos(x) - cos(x) = 0
cos(x)[2sin(x) - 1] = 0
so cos(x) = 0, and/or 2sin(x) - 1 = 0, between 0 and 3pi/2
pi/2, 3pi/2 ............. pi/6, 5pi/6
c) The line is increasing at points in which f'(x) > 0. I'm not sure if there is a better way to do this, but since the tangential lines are zero at (pi/6, pi/2, 5pi/6, 3pi/2), just pick an arbitrary number between each of these intervals (as well as the 0 to pi/6 interval). If the result is positive, that is an increasing interval.
d) every x intercept of f'(x) is a horizontal tangent of the original equation which you found in b). f'(x) represents the rate of change of the original equation, so anywhere the graph of f'(x) is above y=0 the original function is increasing.