Does The Composite Function $h=f(g(x))$ Share The Discontinuity Of ...

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Learn more about Teams Does the composite function $h=f(g(x))$ share the discontinuity of $g(x)$? Ask Question Asked 6 years, 8 months ago Modified 6 years, 8 months ago Viewed 926 times 1 $\begingroup$

For example,

Is the function $h(x)=f(f(x))$ discontinuous at $x=0$ if $f(x)=\frac{1}{x}$ ?

My observation

$h(x)=x$ is continuous at $x=0$

$f(f(x))=\frac{1}{\frac{1}{x}}$ is discontinuous at $x=0$ as $f(x)=\frac{1}{x}$ is discontinuous at $x=0$.

The function $h(x)=x$ and $f(f(x))=\frac{1}{\frac{1}{x}}$ are not exactly the same. Thus, though $h(x)=x$ is continuous, when we say the composite function $f(f(x))=\frac{1}{\frac{1}{x}}=x$ we are already assuming the function is not defined at $x=0$, thus discontinuous at $x=0$.

Is my observation correct ?

Does this applies to similar composite functions, say $h(x)=f(g(x))$ where $f(x)=\frac{1}{x-1}$ and $g(x)=\frac{1}{x-2}$ ?

Similar Post

In a similar problem Discontinuity of composite function , I do not find any explanation rather than few comments arguing the continuity of function $\frac{1}{x}$ does not make sense at $x=0$ and it is a continuous function as the limit exists in its domain. But, check Example 5 clearly states "it is not a continuous function since its domain is not an interval. It has a single point of discontinuity, namely x = 0, and it has an infinite discontinuity there".

• calculus
• continuity

Share Cite Follow asked Apr 10, 2018 at 9:40 SOORAJ SOMANSOORAJ SOMAN 7,92044 gold badges5252 silver badges9595 bronze badges $\endgroup$ 10

• $\begingroup$ If a function $f$ is not defined at $x_0$, it doesn't have the property of being or not being continuous at that point. You can ask for a continuous extension though. $\endgroup$ – Christoph Commented Apr 10, 2018 at 9:44
• $\begingroup$ @Christoph pls check what is being explained in Example 5 in math.mit.edu/~jspeck/18.01_Fall%202014/Supplementary%20notes/… $\endgroup$ – SOORAJ SOMAN Commented Apr 10, 2018 at 9:45
• $\begingroup$ The definition of continuity of a function in those notes includes the domain being an interval. Note that this a rather non-standard definition. $\endgroup$ – Christoph Commented Apr 10, 2018 at 9:48
• $\begingroup$ h (x) is not equal to x but to $\frac {1}{\frac {1}{x}} $ $\endgroup$ – MysteryGuy Commented Apr 10, 2018 at 9:48
• 1 $\begingroup$ Well, clearly, $f(x) = \frac1x$ is not continuous at $x = 0$, that we can agree on. However, I subscribe to a notion that it's not discontinuous either. It's just... not defined. There is definitely an asymptote at $x = 0$, though, and in some circumstances I would call it a pole. I guess you could call that an "infinite discontinuity", if you want. $\endgroup$ – Arthur Commented Apr 10, 2018 at 11:16

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1 Answer 1

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 1 $\begingroup$

In all your examples, you consider functions that are not defined at some point. If $f$ is not defined at $x=0$, then $g\circ f$ is not defined at $x=0$ either and therefore not continuous at $x=0$.

In more generality, if $f$ is defined over all $\Bbb R$, but not continuous, it may very well be that $g\circ f$ is continuous: this happens for example if $g$ is constant - but of course this is not necessary.

Share Cite Follow answered Apr 10, 2018 at 9:49 Arnaud MortierArnaud Mortier 27.5k33 gold badges3636 silver badges8383 bronze badges $\endgroup$ Add a comment |

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2 Discontinuity of composite function $f(g(x))$ where $f(x)=\frac{1}{x^2+x-2}$ and $g(x)=\frac{1}{x-1}$ 0 From the discontinuity of $f(x)$ and $g(x)$, can we directly tell about the discontinuity of $f(g(x))$?

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