SATPREP assignment on domain and range of functions and function composition.

Here are the solutions for the assignment on Domain and Range, followed by the Composite Functions section.

Part 1: Find the Domain and Range

(a) $y = 5x + 3$
* Domain: This is a linear equation (a straight line). There are no restrictions on $x$.
* Domain: All real numbers ($\mathbb{R}$) or $(-\infty, \infty)$
* Range: A non-horizontal straight line covers all possible $y$ values.
* Range: All real numbers ($\mathbb{R}$) or $(-\infty, \infty)$

(b) $y = -7x - 4$
* Domain: Linear equation. No restrictions.
* Domain: All real numbers ($\mathbb{R}$)
* Range: Linear equation. Covers all $y$ values.
* Range: All real numbers ($\mathbb{R}$)

(c) $y = 7$
* Domain: You can plug in any number for $x$, and $y$ will always be 7.
* Domain: All real numbers ($\mathbb{R}$)
* Range: The output is always just 7.
* Range: $\{7\}$

(d) $y = x^2$
* Domain: You can square any real number.
* Domain: All real numbers ($\mathbb{R}$)
* Range: Any number squared is positive or zero. It cannot be negative.
* Range: $y \geq 0$ or $[0, \infty)$

(e) $y = 2x^2 + 1$
* Domain: You can square any number.
* Domain: All real numbers ($\mathbb{R}$)
* Range: Since $x^2 \geq 0$, then $2x^2 \geq 0$. Adding 1 means the smallest value is 1.
* Range: $y \geq 1$ or $[1, \infty)$

(f) $y = x^2 - 2x + 5$
* Domain: Polynomial function.
* Domain: All real numbers ($\mathbb{R}$)
* Range: This is a parabola opening upwards. We find the vertex to get the minimum value.
* Vertex $x = \frac{-b}{2a} = \frac{-(-2)}{2(1)} = 1$.
* Plug $x=1$ back in: $y = (1)^2 - 2(1) + 5 = 1 - 2 + 5 = 4$.
* Range: $y \geq 4$ or $[4, \infty)$

(g) $y = 4 - x^2$
* Domain: Polynomial function.
* Domain: All real numbers ($\mathbb{R}$)
* Range: This is a parabola opening downwards (because of $-x^2$). The maximum value occurs at the vertex ($x=0$).
* Max value: $y = 4 - 0^2 = 4$.
* Range: $y \leq 4$ or $(-\infty, 4]$

(h) $y = -1 + 2x - x^2$
* Domain: Polynomial function.
* Domain: All real numbers ($\mathbb{R}$)
* Range: Parabola opening downwards ($a = -1$). Find the vertex.
* Vertex $x = \frac{-2}{2(-1)} = 1$.
* Plug $x=1$ back in: $y = -1 + 2(1) - (1)^2 = -1 + 2 - 1 = 0$.
* Range: $y \leq 0$ or $(-\infty, 0]$

(i) $y = \sqrt{x + 9}$
* Domain: The inside of a square root must be non-negative.
* $x + 9 \geq 0 \Rightarrow x \geq -9$.
* Domain: $[-9, \infty)$
* Range: The principal square root is always non-negative.
* Range: $y \geq 0$ or $[0, \infty)$

(j) $y = \sqrt{3x - 4}$
* Domain: Inside must be non-negative.
* $3x - 4 \geq 0 \Rightarrow 3x \geq 4 \Rightarrow x \geq \frac{4}{3}$.
* Domain: $[\frac{4}{3}, \infty)$
* Range: Square roots result in non-negative numbers.
* Range: $y \geq 0$ or $[0, \infty)$

(k) $y = \sqrt{x^2 - 4}$
* Domain: Inside must be non-negative.
* $x^2 - 4 \geq 0 \Rightarrow x^2 \geq 4$. This happens when $x$ is "outside" -2 and 2.
* Domain: $(-\infty, -2] \cup [2, \infty)$
* Range: The smallest value inside the root is 0 (when $x=\pm 2$), so the smallest $y$ is 0. As $x$ gets huge, $y$ gets huge.
* Range: $y \geq 0$ or $[0, \infty)$

(l) $y = \sqrt{4 - x^2}$
* Domain: Inside must be non-negative.
* $4 - x^2 \geq 0 \Rightarrow 4 \geq x^2 \Rightarrow x^2 \leq 4$. This happens between -2 and 2.
* Domain: $[-2, 2]$
* Range: The max value inside is 4 (when $x=0$), so max $y = \sqrt{4} = 2$. Min value is 0.
* Range: $[0, 2]$

(m) $y = \frac{1}{7 - x}$
* Domain: Denominator cannot be zero.
* $7 - x \neq 0 \Rightarrow x \neq 7$.
* Domain: All real numbers except 7 ($x \neq 7$)
* Range: A fraction with a constant numerator (1) can never equal 0.
* Range: All real numbers except 0 ($y \neq 0$)

(n) $y = \frac{6}{x + 2}$
* Domain: Denominator cannot be zero.
* $x + 2 \neq 0 \Rightarrow x \neq -2$.
* Domain: All real numbers except -2 ($x \neq -2$)
* Range: Numerator is non-zero constant, so $y$ cannot be 0.
* Range: All real numbers except 0 ($y \neq 0$)

(o) $y = \sqrt{\frac{1}{x - 2}}$
* Domain: Inside root must be $\geq 0$, AND denominator $\neq 0$.
* $\frac{1}{x - 2} \geq 0$. Since numerator is positive, denominator must be positive.
* $x - 2 > 0 \Rightarrow x > 2$.
* Domain: $(2, \infty)$
* Range: As $x \to 2$, $y \to \infty$. As $x \to \infty$, $y \to 0$ (but never reaches it).
* Range: $y > 0$ or $(0, \infty)$

(p) $y = \frac{2}{\sqrt{2x - 5}}$
* Domain: Inside root must be $> 0$ (cannot be 0 because it's in the denominator).
* $2x - 5 > 0 \Rightarrow 2x > 5 \Rightarrow x > 2.5$.
* Domain: $(2.5, \infty)$
* Range: Numerator is positive, denominator is positive. Result is always positive. Can never be 0.
* Range: $y > 0$ or $(0, \infty)$

(q) $y = \frac{x}{\sqrt{x + 1}}$
* Domain: Inside root must be $> 0$ (denominator restriction).
* $x + 1 > 0 \Rightarrow x > -1$.
* Domain: $(-1, \infty)$
* Range: This function crosses the x-axis at $x=0$. It goes to $-\infty$ as $x \to -1$ and $+\infty$ as $x \to \infty$. It is continuous.
* Range: All real numbers ($\mathbb{R}$)

(r) $y = \sqrt{x - 2} + \frac{1}{x - 5}$
* Domain: Root needs $x - 2 \geq 0 \Rightarrow x \geq 2$. Fraction needs $x - 5 \neq 0 \Rightarrow x \neq 5$.
* Combine them: Start at 2, go up, but skip 5.
* Domain: $[2, 5) \cup (5, \infty)$
* Range: *(Note marked with * in image)*
* At $x=2$, $y = 0 + \frac{1}{-3} = -1/3$.
* As $x \to 5$ from left, term $\frac{1}{x-5} \to -\infty$.
* As $x \to 5$ from right, term $\frac{1}{x-5} \to +\infty$.
* Since it goes to both infinities and is continuous in those intervals, it covers everything.
* Range: All real numbers ($\mathbb{R}$)

(s) $y = \frac{1}{1 - \frac{1}{x - 2}}$
* Domain: Two restrictions.
1. Small denominator: $x - 2 \neq 0 \Rightarrow x \neq 2$.
2. Big denominator: $1 - \frac{1}{x - 2} \neq 0 \Rightarrow 1 \neq \frac{1}{x - 2} \Rightarrow x - 2 \neq 1 \Rightarrow x \neq 3$.
* Domain: All real numbers except 2 and 3.
* Range: Let $y = 0$. Then $0 = \frac{1}{\dots}$ which is impossible. So $y \neq 0$.
* Are there other holes? Simplify: $y = \frac{x-2}{x-3}$. Horizontal asymptote is $y=1$. Can $y=1$? $1 = \frac{x-2}{x-3} \Rightarrow x-3=x-2 \Rightarrow -3=-2$ (False). So $y \neq 1$.
* Range: All real numbers except 0 and 1.

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Part 2: Composite Functions

(a) $f(x) = 2x + 4$, $g(x) = 3x - 5$
1. $(f \circ g)(x)$: Plug $g$ into $f$.
$f(3x - 5) = 2(3x - 5) + 4 = 6x - 10 + 4 = 6x - 6$
2. $(g \circ f)(x)$: Plug $f$ into $g$.
$g(2x + 4) = 3(2x + 4) - 5 = 6x + 12 - 5 = 6x + 7$
3. Domain of $(f \circ g)(x)$: Both are lines, no restrictions.
Domain: All real numbers.
4. Range of $(f \circ g)(x)$: It's a line ($6x-6$).
Range: All real numbers.

(b) $f(x) = x^2$, $g(x) = x - 3$
1. $(f \circ g)(x)$: $f(x-3) = (x - 3)^2$
2. $(g \circ f)(x)$: $g(x^2) = x^2 - 3$
3. Domain of $(f \circ g)(x)$: Polynomials have no restrictions.
Domain: All real numbers.
4. Range of $(f \circ g)(x)$: $(x-3)^2$ is a square, so it's always $\geq 0$.
Range: $[0, \infty)$

(c) $f(x) = \sqrt{x + 2}$, $g(x) = 2x - 4$
1. $(f \circ g)(x)$: $f(2x - 4) = \sqrt{(2x - 4) + 2} = \sqrt{2x - 2}$
2. $(g \circ f)(x)$: $g(\sqrt{x + 2}) = 2\sqrt{x + 2} - 4$
3. Domain of $(f \circ g)(x)$: Inside root $\geq 0$.
$2x - 2 \geq 0 \Rightarrow 2x \geq 2 \Rightarrow x \geq 1$.
Domain: $[1, \infty)$
4. Range of $(f \circ g)(x)$: Principal root is $\geq 0$.
Range: $[0, \infty)$

(d) $f(x) = 3x - 4$, $g(x) = \sqrt{x + 2}$
1. $(f \circ g)(x)$: $f(\sqrt{x + 2}) = 3\sqrt{x + 2} - 4$
2. $(g \circ f)(x)$: $g(3x - 4) = \sqrt{(3x - 4) + 2} = \sqrt{3x - 2}$
3. Domain of $(f \circ g)(x)$: Determined by the inner function $g(x)$.
$x + 2 \geq 0 \Rightarrow x \geq -2$.
Domain: $[-2, \infty)$
4. Range of $(f \circ g)(x)$: Smallest value of $\sqrt{x+2}$ is 0.
Min $y = 3(0) - 4 = -4$.
Range: $[-4, \infty)$

(e) $f(x) = \frac{1}{x}$, $g(x) = \frac{1}{x}$
1. $(f \circ g)(x)$: $f(\frac{1}{x}) = \frac{1}{1/x} = x$
2. $(g \circ f)(x)$: Same as above, $x$.
3. Domain of $(f \circ g)(x)$: Inner function $g(x)$ requires $x \neq 0$. Even though the simplified answer is just "$x$", the original input cannot be 0.
Domain: $x \neq 0$
4. Range of $(f \circ g)(x)$: Since $x$ cannot be 0, $y$ cannot be 0.
Range: $y \neq 0$

(f) $f(x) = x^2 + 2$, $g(x) = \sqrt{x - 4}$
1. $(f \circ g)(x)$: $f(\sqrt{x - 4}) = (\sqrt{x - 4})^2 + 2 = (x - 4) + 2 = x - 2$
2. $(g \circ f)(x)$: $g(x^2 + 2) = \sqrt{(x^2 + 2) - 4} = \sqrt{x^2 - 2}$
3. Domain of $(f \circ g)(x)$: Inner function $g(x)$ requires $x - 4 \geq 0 \Rightarrow x \geq 4$.
Domain: $[4, \infty)$
4. Range of $(f \circ g)(x)$: Function is $y = x - 2$ but only for $x \geq 4$.
Min $y = 4 - 2 = 2$.
Range: $[2, \infty)$

(g) $f(x) = \frac{1}{1 - x}$, $g(x) = \sqrt{5 - x}$
1. $(f \circ g)(x)$: $f(\sqrt{5 - x}) = \frac{1}{1 - \sqrt{5 - x}}$
2. $(g \circ f)(x)$: $g(\frac{1}{1 - x}) = \sqrt{5 - \frac{1}{1 - x}}$
3. Domain of $(f \circ g)(x)$:
* Inner root: $5 - x \geq 0 \Rightarrow x \leq 5$.
* Outer denominator: $1 - \sqrt{5 - x} \neq 0 \Rightarrow \sqrt{5 - x} \neq 1 \Rightarrow 5 - x \neq 1 \Rightarrow x \neq 4$.
* Domain: $(-\infty, 4) \cup (4, 5]$
4. Range of $(f \circ g)(x)$: *(Hard problem)*
* As $x \to 4$, denominator $\to 0$, so $y \to \pm \infty$.
* At endpoint $x = 5$, $y = \frac{1}{1 - 0} = 1$.
* As $x \to -\infty$, $\sqrt{5-x} \to \infty$, denominator $\to -\infty$, $y \to 0$.
* Because of the vertical asymptote at $x=4$, the range splits.
* Range: $(-\infty, 0) \cup [1, \infty)$ *(Note: Depending on strictness, usually just identifying it covers most reals except a gap near 0 to 1 is sufficient for high school level unless graphing calculator is used).*

(h) $f(x) = \sqrt{x - 2}$, $g(x) = \frac{3}{x - 9}$
1. $(f \circ g)(x)$: $f(\frac{3}{x - 9}) = \sqrt{\frac{3}{x - 9} - 2}$
2. $(g \circ f)(x)$: $g(\sqrt{x - 2}) = \frac{3}{\sqrt{x - 2} - 9}$
3. Domain of $(f \circ g)(x)$:
* Inner denominator: $x \neq 9$.
* Outer root: $\frac{3}{x - 9} - 2 \geq 0 \Rightarrow \frac{3}{x - 9} \geq 2$.
* Case 1 ($x > 9$): $3 \geq 2(x - 9) \Rightarrow 3 \geq 2x - 18 \Rightarrow 21 \geq 2x \Rightarrow x \leq 10.5$. So, $(9, 10.5]$.
* Case 2 ($x < 9$): LHS is negative, RHS is positive. Never true.
* Domain: $(9, 10.5]$
4. Range of $(f \circ g)(x)$: *(Hard problem)*
* At $x = 10.5$, inside is 0, so $y = 0$.
* As $x \to 9$ (from right), inside $\to \infty$, so $y \to \infty$.
* Range: $[0, \infty)$