0 votes 0 votes Is the function $f(x)=\frac{1}{x^{\frac{1}{3}}}$ continous in the interval [-1 0) ? Calculus engineering-mathematics calculus continuity + – Ayush Upadhyaya asked Jul 17, 2018 Ayush Upadhyaya 660 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments srestha commented Jul 17, 2018 i edited by srestha Jul 17, 2018 reply Follow Share [-1 0) means greater than or equal to -1 and less than 0 so, undefined value isnot there will be continuous https://en.wikipedia.org/wiki/Interval_(mathematics) 0 votes 0 votes srestha commented Jul 17, 2018 reply Follow Share Only 0 is responsible for discontinuity but here 0 point isnot included less than 0 value is there So continuous 0 votes 0 votes Kabir5454 commented May 10, 2022 reply Follow Share The function is continuous as here the interval is given [-1,0). Here, The function is not defined for 0 . but here we have to consider the value x-> $0^{-}$. as x->$0^{-}$, the function $f(x)=\frac{1}{x^{\frac{1}{3}}}$ tends to infinity but still it is continuos and defined . and as well as x->$(-1)^{+}$ the value of the function is tends to -1. so it is countinuos on both the end point of the range. we should consider that the function is not defined for x->$(-1)^{-}$ . 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes The function like this: As x->0 value goes to infinity. Interval given is [-1,0) i.e Closed -1 to open zero. Here as can be seen, it is continues between the given interval. CJ147 answered Aug 9, 2018 CJ147 comment Share Follow See all 0 reply Please log in or register to add a comment.