Functions - Domain and Range; Composition

Sol 1 f is defined for all values of x (since f is a polynomial), so the domain of f is $(-\infty,\infty)$. Since the graph of f is a parabola which opens downward with vertex at $(0,4)$, the set of y-coordinates for the points on the graph of f consists of all y-values with $y\le 4$; so the range of f is the interval $(-\infty,4]$.

Sol 2 f is defined where $9-x\ge 0$ or $9\ge x$, so the domain of f is the interval $(-\infty,9]$. Since $\sqrt{9-x}\ge 0$, $f(x)= 5+\sqrt{9-x}\ge 5$ for any x in the domain of f; so the range of f is contained in $[5,\infty)$. If $y\ge 5$, $f(9-(y-5)^2)=5+\sqrt{9-(9-(y-5)^2)}=5+\sqrt{(y-5)^2} =5+(y-5)=y$ (since $y-5\ge 0$), so the range of f is actually equal to $[5,\infty)$.

Sol 3 f is defined for $x-8\neq 0$ or $x\neq 8$, so the domain of f is given by $(-\infty,8)\cup (8,\infty)$. To find the range of f, we must determine for which y-values the equation $y=\frac{5x}{x-8}$ has a solution for x. Multiplying both sides of this equation by $x-8$ gives $(x-8)y=5x$ or $xy-8y=5x$, so $yx-5x=8y \Rightarrow (y-5)x=8y \Rightarrow x=\frac{8y}{5-y}$. Therefore the equation $y=\frac{5x}{x-8}$ has a solution for x iff $5-y\neq 0$ or $y\neq 5$, so the range of f is given by $(-\infty,5)\cup (5,\infty)$.

Sol 4 f is defined wherever $36-x^2\ge 0$ (so the square root is defined) and $36-x^2 \neq 0$ (so the fraction is defined). Solving the inequality $36-x^2>0$ or $(6-x)(6+x)>0$ gives $-6<x<6$, so the domain of f is $(-6,6)$.

Sol 5 $(f \circ g)(x)=f(\sqrt{2x-3})=(\sqrt{2x-3})^2+4=2x-3+4=2x+1$, while $(g \circ f)(x)=g(x^2+4)=\sqrt{2(x^2+4)-3}=\sqrt{2x^2+5}$.

Sol 6 We can let $g(x)=x^2+6x-4$ and $f(x)=x^5$, for example.

Sol 7 Since $(g \circ f)(x)=x$, $g(f(x))=2(f(x))-9=x$ and therefore $2(f(x))=x+9$ and $f(x)=1/2(x+9)$.

Sol 8 $f$ is defined where $25-x^2\ge0$, so $25\ge x^2$ gives $x^2\le 25$. Taking the nonnegative square root of both sides gives $\vert x\vert\le 5$ or $-5\le x\le 5$. Therefore $[-5,5]$ is the domain of $f$.

Sol 9 $f$ is defined where

$$\frac{x^2-5x}{x^2-9}\ge 0$$

, so factoring gives the inequality

$$\frac{x(x-5)}{(x-3)(x+3)}\ge 0$$

.

Marking 0,5,3, and -3 on a number line and using the facts that all factors have odd exponents and that $x=1\Rightarrow\frac{x^2-5x}{x^2-9}=1/2>0$,

we get the following sign chart for $\frac{x^2-5x}{x^2-9}$:

Therefore the domain of $f$ is given by $(-\infty,-3)\cup[0,3)\cup[5,\infty)$

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