Question

Let f open parentheses x close parentheses equals ln open parentheses x squared close parentheses space space space a n d space space space g open parentheses x close parentheses equals square root of e to the power of 3 x end exponent end root. Find f o g open parentheses x close parentheses and its domain?

Answer 1

Answer:

$(f\ o\ g)(x) = 3x$

$-\infty < x < \infty$

Step-by-step explanation:

See comment for right presentation of question

Given

$f(x) = ln(x^2)$

$g(x)=\sqrt{e^{3x}}$

Solving (a): (f o g)(x)

This is calculated as:

$(f\ o\ g)(x) = f(g(x))$

We have:

$f(x) = ln(x^2)$

$f(g(x)) = \ln((g(x))^2)$

Substitute: $g(x)=\sqrt{e^{3x}}$

$f(g(x)) = \ln(\sqrt{e^{3x}})^2$

Evaluate the square

$f(g(x)) = \ln(e^{3x})$

Using laws of natural logarithm:

$\ln(e^{ax}) = ax$

So:

$f(g(x)) = \ln(e^{3x})$

$f(g(x)) = 3x$

Hence:

$(f\ o\ g)(x) = 3x$

Solving (b): The domain

We have:

$f(g(x)) = 3x$

The above function has does not have any undefined points and domain constraints.

Hence, the domain is: $-\infty < x < \infty$