text: given the first derivative of the function f'(x) how many critical values does it have on the open interval... the first derivateive of the function f is given by f'(x)=((cos^2x)/x)-(1/5). how many critical values does f have on the open interval (0,10)?  ok, i would appreciate any help. i know there are at least four critical points, but my question is this; one of the points on the grapph of f'(x) is tangent to the x-axis. i know that a critical point on the graph of f'(x) occurs when f'(x)=0 or f'(x) is undefined. if the the grapg of f'(x) is tangent to the x-axis, is the point to which it is tangent, critical? i’d suggest you another approach: (cosx)^2=0.5*(1+cos(2x)); if y’=0 then 1+cos(2x)=0.4*x or cos(2x)=0.4*x-1 (the x=0 is outside of our interval). now look at the graphs of y1=cos(2x) and that of y2=0.4x-1; |cos(2x)|<=1; ‘start’ point of  y2 is [0,-1], y2 is rather ‘slow-up’ and y2 must be <= than 1, ie 0.4*x-1<=1 or x<=5; thus our interval is not (0,10) any longer, it is (0,5];  5=1.59*pi; 1.59 is the # of periods of y1 in 5, because 1 period of y1 is 2x=2*pi and x=pi. thus there are only 3 critical points in the interval (0,10)!