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