document.write( "Question 239899: State the domain and range of each function....
\n" ); document.write( "g(x)=ln(2x-1)
\n" ); document.write( "f(x)=1-e^x \n" ); document.write( "

Algebra.Com's Answer #175988 by jsmallt9(3758) $\"\"$ $\"About$

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The keys to these problems are:

Exponents can be any Real number: zero, positive, negative, fractions, irrational, etc. And since logarithms are exponents they too can be any Real number.
The result of raising a positive number to a power will always be positive, no matter what the exponent is! (Remember that zero exponents result in 1's and negative exponents do not result in negative values. Negative exponents mean \"reciprocal of\" and the reciprocal of a positive is also positive.) And since the argument of a logarithm is the result of raising a positive number to a power, the argument must be positive.

\n" ); document.write( "Domain of g(x).
\n" ); document.write( "Since the x is in the argument of a logarithm, the domain must make the argument of the logarithm positive:
\n" ); document.write( " $\"2x-1+%3E+0\"$
\n" ); document.write( "Solving this we get:
\n" ); document.write( " $\"2x+%3E+1\"$
\n" ); document.write( " $\"x+%3E+1%2F2\"$ which is the domain of g(x).

\n" ); document.write( "Range of g(x).
\n" ); document.write( "Since g(x) is a logarithm, the range is all Real numbers.

\n" ); document.write( "Domain of f(x).
\n" ); document.write( "Since x is the exponent, the domain is all Real numbers.

\n" ); document.write( "Range of f(x).
\n" ); document.write( "Since $\"e%5Ex\"$ can be any positive number, from the tiniest fraction (for large negative x's) to the infinitely large (for large positive x's), we just have to figure out what values $\"1-e%5Ex\"$ will be. When $\"e%5Ex\"$ is a tiny fraction, f(x) will be a tiny bit less than 1. When $\"e%5Ex\"$ is infinitely large, f(x) will be an infinitely large negative number. So the range of f(x) is all Real numbers less than 1. \n" ); document.write( "

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