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"A function f(x) has domain $\{x \in \mathbb{R}\;|\;x \ge -4\}$ and range $\{y \in \mathbb{R}\;|\;y \lt -1\}$ Determine the domain and range of each function without graphing."

I was given the $2$ functions $y=f(-x)$ and $y=-2f(-x+5) + 1$. For the first function, I understand why the the domain is $x \lt 4$ (everything is flipped because of $-1$), but I don't understand why it doesn't say that $x \le 4$.

For the second function, I don't understand why the range doesn't say $y$ $\gt 3$ . I also don't understand why the domain isn't $x \le -1$. Hopefully this post made sense.

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    $\begingroup$ Please check the first sentence in the last paragraph: "is less is greater than $1$" does not make sense. Also, please read this tutorial on how to typeset mathematics on this site. Using the proper mathematical symbolism would make your post easier to read. $\endgroup$ Commented Oct 2, 2016 at 21:51
  • $\begingroup$ I don't understand how the LaTeX works. I typed y \ge 3 but it didn't work. $\endgroup$
    – Hamze
    Commented Oct 2, 2016 at 22:39
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    $\begingroup$ See MathJax basic tutorial and quick reference. You need to enclose it in $ symbols, like $y \ge 3$. $\endgroup$
    – dxiv
    Commented Oct 2, 2016 at 22:44

1 Answer 1

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For the second function, let $g(x) = -2 f(-x+5) + 1$.

The argument passed to $f$ is $-x + 5$, and it must be in the domain of $f$, so:

$$-x + 5 \ge -4 \quad \iff \quad x \le 9$$

The range of values for $g(x)$ follows from:

$$f(-x + 5) \lt -1$$ $$-2 f(-x + 5) \gt 2$$ $$g(x) = -2 f(-x + 5) + 1 \gt 3$$

From the above, the domain of $g$ is $\{x \in \mathbb{R}\;|\; x \le 9\}$ and its range is $\{y \in \mathbb{R}\;|\; y \gt 3\}$.

The first function can be worked out the same way, and it's obviously simpler.

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  • $\begingroup$ $\{x \in \mathbb{R}\;|\;x \lt -1\}$ and $\{y \in \mathbb{R}\;|\;y \lt 3\}$ was the textbook's answer. Would you say the textbook's answer is incorrect? $\endgroup$
    – Hamze
    Commented Oct 2, 2016 at 22:53
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    $\begingroup$ @Hamze That's either wrong, or maybe refers to a problem other than the one you posted. Simply by inspection it's easy to see that $f$ is always negative, so $-2 f(-x+5)$ is always positive, therefore the range of $y = -2 f(-x+5) + 1$ can not include any negative numbers. $\endgroup$
    – dxiv
    Commented Oct 2, 2016 at 23:01
  • $\begingroup$ I don't understand, how do you know f is always negative? $\endgroup$
    – Hamze
    Commented Oct 2, 2016 at 23:06
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    $\begingroup$ The domain is $x \ge -4$ and the range is $y \lt -1$ so $f(x) < -1 \lt 0$ for $\forall x \ge -4$. $\endgroup$
    – dxiv
    Commented Oct 2, 2016 at 23:08
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    $\begingroup$ From the answer: "The argument passed to $f$ is $−x+5$, and it must be in the domain of $f$". Stated another way, $f( \cdot )$ is defined for $\cdot \ge -4$. Replace the $\cdot$ with $-x+5$ and the inequality follows as written. $\endgroup$
    – dxiv
    Commented Oct 2, 2016 at 23:52

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