Draw the graph of `y=sin^(-1)("log"_(e)x)`. Also find the point of inflection.

Explanation

We have `y=f(x) =sin^(-1)("log"_(e)x)`
`Heref(x) is defined if-1le" log"_(e)xle1`
`or " "e^(-1)lexle e`
`f(x) = sin^(-1)("log"_(e)x)=0`
`therefore" log"_(e)x=0`
`therefore" "x=1`
So the graph meets the x-axis at (1,0).
`f'(x)=1/(xsqrt(1-(log_(e)x))^(2))gt0," hence f(x) is invreasing".`
`f'(x)=(sqrt(1-(log_(e)x)^(2))-(log_(e)x)/(sqrt((1-(log_(e)x)^(2)))))/(x^(2)(1-(log_(e)x)^(2)))`
`f'(x)=sqrt(1-(log_(e)x)^(2))-(log_(e)x)/(sqrt((1-(log_(e)x)^(2))))=0`
`therefore" "(log_(e)x)^(2)+log_(e)x-1=0`
`therefore" "log_(e)x=(-1+sqrt5)/2`
`therefore" "x=e^((-1+sqrt5)/2)`
Hence the graph of the function is as shown in the following figure.

In the figure, the graph has his point of infection at point A.