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

Answer

7

Explanation

Integrating both sides w.r.t `x`
`(dy)/(dx)=3x^(2)-4x+c`
at `x=1 (dy)/(dx)=0impliesc=1`
`implies(dy)/(dx)=3x^(2)-4x+1`…………(1)
Integeratig both sides w.r.t `x`
`y=x^(3)-2x^(2)+x+c_(1)`
at `x=1, y=5impliesc_(1)=5`
`impliesy=x^(2)-2x^(2)+x+5`
from equation (1) we get the critical points `x=1/3,1`
`f(1)=5`
`f(0)=5,f(2)=7 1(1/3)=112/27`
Hence the global maximum value `=7`