Let `f(x)` have a point of inflection at `x=1` and let `f^(")(x)=x.` If `f^(')(1)=0,` then `f(x)` is equal to

Options

A. `x^(4)-6x^(2)+8x`
B. `x^(4)-6x^(2)-x`
C. `1/24 (x^(4)-6x^(2)+8x)`
D. `1/12(x^(4)-6x^(2)+8x)`

Answer

C

Explanation

`f^(")(x)=x`
`f^(")=(x^(2))/2+A`
`:'f^(")(1)=0`, as `x=1` is point of infection
`A=-1/2`
`f^(")(x)=(x^(2))/2-1/2`
`f^(')(x)=(x^(3))/6-1/2x+B`
`f^(')(1)=0impliesB=1/2-1/6=1/3`
`:.f^(')(x)=(x^(3))/6-x/2+1/3`
`:.f(x)=(x^(4))/24-1/4x^(2)+1/3x+C`
`f(0)=0impliesc=0`
`:.f(x)=(x^(4))/24-1/4x^(2)+1/3x`