Homework Soursewhat are the critical numbers of gt l 3t4 l Im not sure how
what are the critical numbers of g(t)= l 3t-4 l? I\'m not sure how to approach finding the critical numbers in the case of absolute values. could someone please explain this to me?
Solution
g(t)=3t-4, 3t-4>=0 -(3t-4), 3t-4<0 = 3t-4, t>=4/3 4-3t, t<4/3 The critical points are points in domain where derivative is undefined or derivative is 0. If t different than 4/3, the derivative exists and is nonzero The derivative from the left at 4/3 is (3t-4)\'=3 the derivative from the right at 4/3 is (4-3t)\'=-3 The lateral derivatives are not equal, so the derivative at 4/3 does not exist. This means 4/3 is a critical point and is the only one. Basically what happens at 4/3 is that it\'s a sharp turn in the direction of the function, which looks like a V. The same thing can be said about f(x)=|x| which has the graph like a V and 0 is a critical point for |x\\.
