1

Given:

$a\left(x\right)=e^x$

$b\left(x\right)=\left|x+2\right|$

$c\left(x\right)=\frac{\left(x-2\right)}{\left(x+1\right)}$

What is:

  1. $\left(\frac{a\cdot b}{c}\right)\left(3\right)$

  2. The domain of $\left(a^{-1}○a^{-1}\right)\left(x\right)$?

  3. The range of $\left(b○a\right)\left(x\right)$?

I have gotten to the point of being able to solve for function composition. For example, I was able to figure out what $\left(b○a\right)\left(x\right)$ was; however, I do not know how to express the range. Anything is helpful.

So far my answers are:

  1. Domain: (1, $\infty $) Range: ℝ

  2. Domain: ℝ Range: (2, $\infty $)

Jeffrey Liu
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2 Answers2

1

In general, range is harder to figure out than domain. For domain, you determine which elements lead to a problem, like dividing by zero or taking the log of a nonpositive.

For example, $$a^{-1}\circ a^{-1}(x)=\ln \ln x$$ For this to make sense, you need $\ln x >0$, or else the outer log doesn't work. This arises when $x>1$.

The range of $|e^x+2|$ can be found in pieces. $e^x$ has range $(0,+\infty)$. Hence $e^x+2$ has range $(2,+\infty)$. Taking the absolute value of any of these numbers leaves them all unchanged, so the final range is $(2,+\infty)$.

vadim123
  • 82,796
  • Thank you. After reading your answer and solving the problem once again, the domain for the second problem would be (1,+∞)? – Jeffrey Liu Dec 23 '14 at 22:47
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Hints for the second and third questions:

  • By definition, $a^{-1}$ is the function $\ln$ (natural log). Its domain is $(0,\infty)$ and its range is $\mathbb{R}$. Additionally, $\ln(x)\in (0,\infty)$ when $x\in(1,\infty)$.

  • The range of $a(x)=e^x$ is $(0,\infty)$. As $y$ ranges over all positive numbers, what does the quantity $b(y)=|y+2|$ range over?

someone you know
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