Finding domain for when composite function is continuous

In summary, the conversation is about finding where the function ##h(x) = \ln(x^2)## is continuous on its entire domain. The reasoning behind this is that since the natural log is defined for positive values of x, the argument of the function, ##x^2##, must also be positive. Therefore, it can be determined that the function is not continuous at ##x=0##. The individual also asks for help in proving this by solving the equation. It is mentioned that the function is differentiable at all points except for ##x=0##. It is clarified that the function is actually written as ##\ln(x^2)## with a lowercase "l" instead of an uppercase "I".
  • #1
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
I am trying to find where ##h(x) =In{x^2}## is continuous on it's entire domain.

My reasoning is since natural log is defined for ##x > 0##, then the argument ##x^2## should be positive, ##x^2 > 0##, we can see without solving this equation that ##x ≠ 0## for this equation to be true, however, does someone please know how we could prove this by solving that equation for x?

My working is
##x > 0## (Taking square root of both sides of the equation)

Many thanks!
 
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  • #2
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

I am trying to find where ##h(x) =In{x^2}## is continuous on it's entire domain.

My reasoning is since natural log is defined for ##x > 0##, then the argument ##x^2## should be positive, ##x^2 > 0##, we can see without solving this equation that ##x ≠ 0## for this equation to be true, however, does someone please know how we could prove this by solving that equation for x?

My working is
##x > 0## (Taking square root of both sides of the equation)

Many thanks!
It is neither defined nor continuous at ##x=0.## It is continuous everywhere else. What do you use to prove continuity? E.g. it is differentiable at ##x\neq 0## and therewith continuous. Or you use a definition for continuity. There are a few, so which one do you use?
 
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  • #3
ChiralSuperfields said:
I am trying to find where ##h(x) =In{x^2}## is continuous on it's entire domain.
There is no "##In()## function; i.e., starting with uppercase i. It's ##\ln()##, with a lowercase letter l (ell), short for logarithmus naturalis.
 
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  • #4
Note that [itex]\ln x^2 = 2\ln |x|[/itex].
 
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Likes member 731016 and SammyS

Related to Finding domain for when composite function is continuous

What is a composite function?

A composite function is a function that is formed by applying one function to the results of another. If you have two functions, f(x) and g(x), the composite function is written as (f ∘ g)(x) = f(g(x)).

How do you determine the domain of a composite function?

To determine the domain of a composite function (f ∘ g)(x), you need to find the domain of g(x) and then determine where f(g(x)) is defined. Specifically, the domain of (f ∘ g)(x) consists of all x-values that are in the domain of g(x) and for which g(x) lies in the domain of f(x).

What does it mean for a composite function to be continuous?

A composite function (f ∘ g)(x) is continuous if both f and g are continuous functions and the composition does not introduce any discontinuities. This means that for every point c in the domain of (f ∘ g)(x), the limit of (f ∘ g)(x) as x approaches c equals (f ∘ g)(c).

How do you check the continuity of a composite function?

To check the continuity of a composite function (f ∘ g)(x), you need to verify that both f and g are continuous on their respective domains. Additionally, you must ensure that the composition does not introduce any points of discontinuity, which involves checking that g(x) lies within the domain of f for all x in the domain of g.

What are common pitfalls when determining the domain and continuity of composite functions?

Common pitfalls include neglecting to consider the domain restrictions of both functions involved in the composition. Another common mistake is assuming that if f and g are continuous individually, their composition will automatically be continuous without verifying the overlap of their domains. It is also important to check for any values that might cause division by zero or other undefined operations within the composite function.

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