For the function `f(x)=x^4(12(log)_e x-7),` the point (1,7) is the point of inflection. `x=e^(1/3)` is the point of minima the graph is concave downwards in (0,1) the graph is concave upwards in `(1,oo)`

Options

A. the point (1,-7) is the point of minima
B. x=`e^(1//3` is the point of minima
C. the graph is concave downwards in (0,1)
D. the graph is concave upwards in `(1,oo)`

Answer

1,2,3,4

Explanation

`y=f(x)=x^(4)(12log_(e)x-7),xgt0`
`therefore(dy)/(dx)=16x^(3)(3log_(e)x-1)and (d^(2)y)/(dx^(2)) 16x^(2)(9 log_(e)x)`
`(dy)/(dx)=0 rarr x=e^(1//3)`
`At x =-e^(1//3), (d^(2)y)/(dx^(2))gt0 `,hence `x =e^(1//3)` is point of minima
Also for `0ltxlt1,(d^(2)y)/(dx^(2))lt0` and for `xgt1,(d^(2)y)/(dx^(2))gt0`
Hence x =1 is point of inflection and for `ltxlt1` graph is concave downward and for `xgt1` graph is concave upward.