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…

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For a certain curve y=f(x) satisfying (d^(2)y)/(dx^(2))=6x-4, f(x) has a local minimum value 5 when x=1, Find the equation of the curve and also the gobal maximum and global minimum values of f(x) given that 0lexle2.

For a certain curve y=f(x) satisfying (d^(2)y)/(dx^(2))=6x-4, f(x) has a local minimum value 5 when x=1, Find the equation of the curve and also the gobal maximum and global minimum values of f(x) given that 0lexle2.

The curve y =f (x) satisfies (d^(2) y)/(dx ^(2))=6x-4 and f (x) has a local minimum vlaue 5 when x=1. Then f^(prime)(0) is equal to :

The curve y =f (x) satisfies (d^(2) y)/(dx ^(2))=6x-4 and f (x) has a local minimum vlaue 5 when x=1. Then f^(prime)(0) is equal to :

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For certain curve y=f(x) satisfying (d^(2)y)/(dx^(2))=6x-4, f(x) has local minimum value 5 when x=1 Global maximum value of y=f(x) for x in [0,2] is