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

Options

A. 5
B. 7
C. 8
D. 9

Answer

B

Explanation

Integrating `(d^(2)y)/(dx^(2))=6x-4`, we get `(dy)/(dx) = 3x^(2)-4x+A`
When `x=1, (dy)/(dx)=0`, so that `A=1`,Hence,
`(dy)/(dx) = 3x^(2)-4x+1`
Integrating, we get `y=x^(3)-2x^(2)+x+5`.
From equation (1), we get the cricitical points `x=1//3, x=1`.
At the critical point `x=1/3, (d^(2)y)/(dx^(2))` is negative
Therefore, at `x=1//3, y` has a local maximum.
At `x=1, (d^(2)y)/(dx^(2))` is positive.
Therefore, at `x=1, y` has a local minimum.
Also, `f(1) =5, f(1/3)=139/7, f(0) =5, f(2)=7`
Hence, the global maximum value =7 and the global minimum value =5