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 parenthese…

Question

asked Feb 21, 2022 217k views

1 Answer

James Coote · Answer 1 · 2022-02-27T07:34:06+0000

Answer:

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

$-\infty < x < \infty$

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:
$-\infty < x < \infty$