1.

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

Answer»

Solution :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.


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