anyone finding the area....pls upload a graph too...
1ne one?
14 Answers
1See, using the property of mod, we can say that the graph changes its concavity at x=-1/2 in the max function and x= 1/2 in the min function.
to find area, break the limits at these points. And the graph will be that of x2+x and x2-x in the respective interval.
1I dunno. I'm still trying. Just said what crossed my mind at the first glance.
1if it would have had been that simple i wouldnt have uploaded it...
24Surely its a very good question..
but I am struck in the middle of ques....
Maybe Nishant sir or Anant sir can help u out in a better way
1there is only one function but check the limits...the limits are diff.
24I ma getting g(x) as ..
Plzz tell if its correct or not...I need some expert's opinion
66Eureka, your g(x) is not correct.
You see, that for -2 ≤ x < 0, g(x) is same as max f(t) for -2 ≤ t ≤ x. Notice that for t < 0, f(t) is an upward opening parabola having roots at t = -1 and 0. So the maximum of f(t) always occurs at t = -2. So g(x) = 2 for -2 ≤ x < 0.
Once you go to the positive side, f(t) is an upward opening parabola with roots at t =0, 1. So g(x) = f (x) as long as 0 ≤ x ≤ 1/2, after which the minimum is at x=1/2. So g(x) gets latched to f(1/2). As such
g(x) = 2 for -2 ≤ x < 0
= x2 - x for 0 ≤ x ≤ 1/2
= -1/4 for 1/2 < x ≤ 3
So the required area = 113/24 sq units