What is the order of operations when performing transformations of functions?

Question

If the domain of $f(x)$ is $[-10, 10]$, the range is $[-8, 8]$, and $h(x) = -2f\bigl(2(x - 3)\bigr) + 4$, what are is the domain and range of $h(x)$?

I am confused on the order in which I should perform the transformations. My teacher had said that reflections should be performed first, but then when I search for answers online they differ. How would I solve for the domain and range using the transformations in the correct order?

Answer 1

There is not one single order, as some transformations can be swapped. You need to be careful that the order of transformations you choose gives the correct result (either in your mind or by detailed computations). For example using detailed computations, to obtain $h(x) = -2f(2(x - 3)) + 4$ from $f$, you can do (note that the terminology varies from book to book):

1. a horizontal shrink of factor $1/2$: you obtain $g_1(x)=f(2x)$
2. a horizontal shift of $3$ units to the right: you obtain $g_2(x)=g_1(x-3)=f(2(x-3))$
3. a vertical stretch of factor $2$: you obtain $g_3(x)=2g_2(x)=2f(2(x-3))$
4. a vertical flip (that is, a reflection about the $x$-axis): you obtain $g_4(x)=-g_3(x)=-2f(2(x-3))$
5. a vertical shift of $4$ units up: you obtain $g_5(x)=g_4(x)+4=-2f(2(x-3))+4=h(x)$, as wanted.

Note that the two first transformations cannot be swapped as you would obtain $f(2x-3)$, but the third and fourth can be swapped to give the same result.

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Now, for the domain, you don't need to care about the order of transformations: $h(x)$ is defined if and only if $f(2(x-3))$ is defined, which means

$$ -10 \le 2(x-3) \le 10 \iff -5 \le x-3 \le 5 \iff -2 \le x \le 8 $$

so the domain of $h$ is $[-2,8]$.

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The computations for the range are similar: since the range of $f$ is $[-8,8]$, we know that $-8\le f(2(x-3))\le 8$. Then $-16 \le -2f(2(x-3))\le 16$, so $-12\le h(x)\le 20$.

Conversely (*), if $-12\le y\le 20$, then $-16 \le y-4\le 16$ so $-8\le \frac{y-4}{-2}\le 8$. Since the range of $f$ is $[-8,8]$, then there exists $u$ in $[-4,4]$ such that $f(u)=\frac{y-4}{-2}$, so $y=-2f(u)+4=-2f\left(2\left(\frac{u+6}{2}-3\right)\right)=h\left(\frac{u+6}{2}\right)$, so $y$ is in the range of $h$.

(*) I believe most precalculus courses will skip this part as the computations are more complicated.

Answer 2

So, regarding your problem, first look for the independent variable, $x$.

Note it doesn't occur by itself as an argument to $f(x)$. Instead you ahve $f( 2(x-3))$.

Since $[-10,10]$ is the domain of $f(x)$, it only makes since to input those values to $f$.

$2(x-3)\in [-10,10] \implies x \in [-2,8]$. So the domain of $h(x)$ is $[-2,8]$.

The range of $f(x)$ is $[-8,8]$.

$h(x)$ can be rewritten as $h(y)=-2y+4, y=f(x)$.

With that substitution, $y\in [-8,8] \implies h(y)=[-12, 20]$, so the range of $h(x)$ is $[-12,20]$.

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Now consider $h(x)=\frac{1}{3-g(2\sqrt{5-x})^2}$ Where the domain of $g(x)$ is $[-10,10]$.

$2\sqrt{5-x} \in [-10,10] \implies \sqrt{5-x} \in [-5,5]\implies 5-x \in [0,25]\implies -x \in [-5,20]\implies x \in [-20,5]$

So the domain if $h(x)$ is $[-20,5]$

Suppose the range of $g(x) \in [-2,2]$.

$g(x) \in [-2,2] \implies g(x)^2 \in [0,3)\cup (3,4]$

Note $3$ is not allowed since that would make the denominator 0.

Continuing: $g(x)^2 \in [0,3)\cup (3,4] \implies 3-g(x)^2 \in (0,3] \cup [-1,0)\implies \frac{1}{3-g(x)^2}\in (-\infty, -1] \cup [1/3, \infty )$

Gives the range of $h(x)$.