Finding domain for when composite function is continuous

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SUMMARY

The function h(x) = ln{x^2} is continuous for all x except at x = 0. The natural logarithm is defined only for positive arguments, which necessitates that x^2 > 0, leading to the conclusion that x cannot equal zero. To prove continuity, one can demonstrate differentiability for x ≠ 0 or apply a formal definition of continuity. Additionally, it is important to note the correct notation for the natural logarithm, which is ln, not In.

PREREQUISITES
  • Understanding of natural logarithm properties and definitions
  • Knowledge of continuity and differentiability in calculus
  • Familiarity with composite functions
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the definition of continuity in calculus
  • Learn about differentiability and its implications for continuity
  • Explore the properties of logarithmic functions, specifically ln(x)
  • Investigate the behavior of composite functions and their domains
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Students studying calculus, particularly those focusing on continuity and differentiability of functions, as well as educators teaching these concepts in mathematics.

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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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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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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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Note that \ln x^2 = 2\ln |x|.
 
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