For certain curves y= f(x) satisfying `[d^2y]/[dx^2]= 6x-4`, f(x) has local minimum value 5 when x=1. 9. Number of critical point for y=f(x) for x € [0,2] (a) 0 (b)1. c).2 d) 3 10. Global minimum value y = f(x) for x € [0,2] is (a)5 (b)7 (c)8 d) 9 11 Global maximum value of y = f(x) for x € [0,2] is (a) 5 (b) 7 (c) 8 (d) 9

Options

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

Answer

A

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