The global maximum of` f(x) = 1 + 12|x| – 3x^2`? on [-1.4] is equal to

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

Let f(x) = x^50-x^20 STATEMENT-1: Global maximum of f(x) in [0, 1] is 0 STATEMENT-2 : x = 0 is a stationary point of f(x)

A curve y=f(x) satisfies (d^2y)/dx^2=6x-4 and f(x) has local minimum value 5 at x=1 . If a and b be the global maximum and global minimum values of f(x) in interval [0,2] , then ab is equal to…

If f(x+4) = x^(2) - 1 , then f(x) is equal to

If f(x)=x^(3)+3x^(2)+3x , then f(x+1) is equal to

Q.8 If f(x) = |4 x-x^2-3| when x belongs to[0,4] then (A)x = 0 is a global maximum (B) x = 4 is a global maximum (C) x = 2 is a local maximum (D) x = 1 and 3 are global minimum

Let f(x) =x^(3) =x^(2) -x-4 (i) find the possible points of maxima // minima of f(x) x in R (ii) Find the number of critical points of f(x) x in [1,3] (iii) Discuss absolute (global) maxima // minima value of f(x) for x in [-2,2] (iv) Prove that for x in (1,3) the function does not has a Global maximum.

Maximum value of f(x)= (x-2)^(2) . (x-3) +1 is