Calculus-Self Doubt

Question

Is the function $f(x)=\frac{1}{x^{\frac{1}{3}}}$ continous in the interval [-1 0) ?

Comments

[-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)

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Only  0 is responsible for discontinuity

but here 0 point isnot included

less than 0 value is there

So continuous

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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->$(-1)^{-}$ .

Answer 1

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.