Question

the global maximum of f(x) =1+12|x|-3x^2 on [-1,4]is equal to ​

Answer 1

Answer:

Steps to find out the global maxima or minima in [a, b]

Step 1: Find out all the critical points of f(x) in (a, b). Let C1, C2,….Cn be the different critical points.

Step 2: Find the value of the function at these critical points and also at the end points of the domain. Let the values of the function at critical points be f(C1), f(C2)………..f(Cn).

Step 3: Find M1 =max{ f(a), f(C1), f(C2)………..f(Cn), f(b)} and M2= min{ f(a), f(C1), f(C2)………..f(Cn), f(b)}. Now M1is the maximum value of f(x) in [a, b] and M2 is the minimum value of f(x) in [a, b].

Case 2: Global maxima or minima in (a, b):

To find the global maxima and minima in (a, b) step 1 and 2 is same but after that we have to be a bit careful.   After step 1 and 2 find

M1 =max{ f(C1), f(C2)………..f(Cn)} and M2= min{f(C1), f(C2)………..f(Cn)}.

Now if x approaches a- or if x approaches b- , the limit of f(x) > M1 or its limit f(x) < M1 would not have global maximum (or global minimum) in (a, b) but if as x approaches a- and x approaches b- , lim f(x) < M1 and lim f(x) > M2, then M1 and M2 will respectively be the global maximum and global minimum of f(x) in (a,b).

Result: If f(x) is a continuous function on a closed bounded interval [a,b], then f(x) will have a global maximum and a global minimum on [a,b]. On the other hand, if the interval is not bounded or closed, then there is no guarantee that a continuous function f(x) will have global extrema.

Example: f(x) =x2 does not have a global maximum on the interval [0, ∞), the function f(x) =-1/x does not have a global minimum on the interval (0, 1).

Result: If f(x) is differentiable on the interval I, then every global extremum is a local extremum or an endpoint extremum.

Step-by-step explanation: