1. Given $f(x) = x^2 + 3x - 4$: 1. a) Find $f(0)$: Substitute $x=0$ into $f(x)$: $$f(0) = 0^2 + 3(0) - 4 = -4$$ 1. b) Find $f(2)$: Substitute $x=2$: $$f(2) = 2^2 + 3(2) - 4 = 4 + 6 - 4 = 6$$ 1. c) Find $f(h)$: Substitute $x=h$: $$f(h) = h^2 + 3h - 4$$ 1. d) Find $f(2x)$: Substitute $x$ by $2x$: $$f(2x) = (2x)^2 + 3(2x) - 4 = 4x^2 + 6x - 4$$ 1. e) Find $f(x) + f(h)$: Add the expressions: $$f(x) + f(h) = (x^2 + 3x - 4) + (h^2 + 3h - 4) = x^2 + 3x - 4 + h^2 + 3h -4 = x^2 + h^2 + 3x + 3h - 8$$ --- 2. Given $f(x) = \sqrt{x} + 1$ and $g(x) = \sqrt{x} - 4$: First, note the domain of $f$ and $g$ is $x \geq 0$ because of the square roots. 2. a) $(f+g)(x) = f(x) + g(x) = (\sqrt{x} + 1) + (\sqrt{x} - 4) = 2\sqrt{x} - 3$ - Domain: $x \geq 0$ - Range: Since $\sqrt{x} \geq 0$, minimum of $2\sqrt{x} - 3$ is at $x=0$ which is $-3$, and for large $x$ it tends to infinity, so range is $[-3, \infty)$ 2. b) $(f-g)(x) = f(x) - g(x) = (\sqrt{x} + 1) - (\sqrt{x} - 4) = 5$ - Domain: $x \geq 0$ - Range: Constant function $5$, so range is $ {5}$ 2. c) $(f \cdot g)(x) = f(x)g(x) = (\sqrt{x} + 1)(\sqrt{x} - 4) = x - 4\sqrt{x} + \sqrt{x} - 4 = x - 3\sqrt{x} - 4$ - Domain: $x \geq 0$ - Range: Analyze Minimum occurs by setting derivative zero, but for simplicity note: As $x \to 0$, value $ o -4$; As $x$ grows large, term $x$ dominates, so tends to infinity. So range is $[-4, \infty)$ 2. d) $(f/g)(x) = \frac{f(x)}{g(x)}= \frac{\sqrt{x} + 1}{\sqrt{x} - 4}$ - Domain: $x \geq 0$ and denominator $\neq 0 \implies \sqrt{x} \neq 4 \implies x \neq 16$ - Range: all real numbers except value at $x=16$ is undefined; the fraction can take most real values. Exact range is $(-\infty, \infty)$ excluding the value where denominator zero. --- 3. Given $f(x) = \sqrt{x}$ and $g(x) = x^2 - 1$: 3. a) Find $f \circ f$ meaning $f(f(x))$: $$f(f(x)) = f(\sqrt{x}) = \sqrt{\sqrt{x}} = x^{1/4}$$ - Domain: since $f(x)=\sqrt{x}$ requires $x \geq 0$, then applying $f$ twice requires $x \geq 0$ - Range: $x^{1/4} \geq 0$, so range is $[0, \infty)$ 3. b) Find $g \circ g$: $$g(g(x)) = g(x^2 - 1) = (x^2 - 1)^2 - 1 = (x^2 - 1)^2 - 1 = x^4 - 2x^2 + 1 - 1 = x^4 - 2x^2$$ - Domain: all real numbers because polynomials are defined everywhere - Range: since it is quartic, range is $[ -1, )$? Actually the minimum must be checked. Let's check the minimum by derivative: $$h(x) = x^4 - 2x^2$$ $$h'(x) = 4x^3 - 4x = 4x(x^2 -1) = 0$$ Critical points at $x=0, x= 1, x= -1$ Evaluate: $h(0) = 0, h(1) = 1 - 2 = -1, h(-1) = 1 - 2 = -1$ So minimum value is $-1$ and range is $[-1, )$ 3. c) Find $f \circ g$: $$f(g(x)) = f(x^2 -1) = \sqrt{x^2 - 1}$$ - Domain: The radicand $x^2 -1 \geq 0 \implies x^2 \geq 1 \implies x \leq -1$ or $x \geq 1$ - Range: Since square root output is $\geq 0$ and input $x^2 -1$ will be at least $0$, range is $[0, \infty)$ 3. d) Find $g \circ f$: $$g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 - 1 = x - 1$$ - Domain: $f(x)=\sqrt{x}$ requires $x \geq 0$ - Range: Since $x-1$ with $x \geq 0$, minimum at $0-1=-1$, range is $[-1, \infty)$