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

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

Maximum value of f(x)=2x^(3)-3x^(2)-12x+6 is

Maximum value of f(x) =2x^(3)-9x^(2)+12x-2 is

The minimum value of function f(x)=3x^4-8x^3+12x^2-48x+25 on [0,3] is equal to

The global maximum value of f(x0=log_(10)(4x^(3)-12x^(2)+11x-3),x in[2,3] is -(3)/(2)log_(10)3(b)1+log_(10)3log_(10)3(d)(3)/(2)log_(10)3

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)=|x-1|+|x-2|-|x-3|, then (a) maximum value of f(x) is 3 if x in[2,3] (b) maximum value of f(x) is -3 if x in[1,3] (c) maximum value of f(x) is -3 if x in[1,2] (d) maximum value of f(x) is 3 if x in[1,3]

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)

Let a and b respectively be the points of local maximum and local minimum of the function f (x) = 2x^3 - 3x^2 - 12x. If A is the total area of the region bounded by y = f (x), the x-axis and the lines x = a and x = b, then 4 A is equal to __________.

Let f(x)=2x^(3)=9x^(2)+12x+6. Discuss the global maxima and minima of f(x) in [0,2] and (1,3) and,hence,find the range of f(x) for corresponding intervals.