1. Problem 10: Find the range of the given functions with specified domains. 2. For each function, the range is the set of all possible output values $f(x)$ when $x$ varies over the domain. 3. Function 1: $f(x) = 5x + 1$, domain $-5 \leq x \leq 5$. - Calculate $f(-5) = 5(-5) + 1 = -25 + 1 = -24$. - Calculate $f(5) = 5(5) + 1 = 25 + 1 = 26$. - Since $f(x)$ is linear and increasing, range is $[-24, 26]$. 4. Function 2: $f(x) = 4 - 2x$, domain $\{ -1, 0, 1, 2, 3, 4 \}$. - Calculate $f(-1) = 4 - 2(-1) = 4 + 2 = 6$. - Calculate $f(4) = 4 - 2(4) = 4 - 8 = -4$. - Evaluate all values: $6, 4, 2, 0, -2, -4$. - Range is $\{ -4, -2, 0, 2, 4, 6 \}$. 5. Function 3: $f(x) = x^2$, domain $0 \leq x \leq 10$. - Calculate $f(0) = 0^2 = 0$. - Calculate $f(10) = 10^2 = 100$. - Since $x^2$ is increasing on $[0,10]$, range is $[0, 100]$. 6. Function 4: $f(x) = 250 - 12.5x$, domain $0 \leq x \leq 10$. - Calculate $f(0) = 250 - 12.5(0) = 250$. - Calculate $f(10) = 250 - 12.5(10) = 250 - 125 = 125$. - Since $f(x)$ is linear and decreasing, range is $[125, 250]$. 7. Problem 11: Given $f(x) = 4x - 2$ and $g(x) = x^2 - 8x + 15$, find values at $x = -2$. 8. a) Calculate $f(-2) = 4(-2) - 2 = -8 - 2 = -10$. 9. b) Calculate $g(-2) = (-2)^2 - 8(-2) + 15 = 4 + 16 + 15 = 35$. 10. Problem 12: Given $f(x) = 128x - 15$, domain $-3 < x < 15$. 11. a) Calculate $f(\frac{3}{2}) = 128 \times \frac{3}{2} - 15 = 192 - 15 = 177$. 12. b) Find the range of $f$. - Calculate $f(-3) = 128(-3) - 15 = -384 - 15 = -399$ (not included since $x > -3$). - Calculate $f(15) = 128(15) - 15 = 1920 - 15 = 1905$ (not included since $x < 15$). - Range is $(-399, 1905)$. 13. c) Find $a$ such that $f(a) = 1162.6$. - Solve $128a - 15 = 1162.6$. - Add 15: $128a = 1177.6$. - Divide: $a = \frac{1177.6}{128}$. - Simplify: $a = 9.2$. Final answers: - Problem 10 ranges: $[-24, 26]$, $\{ -4, -2, 0, 2, 4, 6 \}$, $[0, 100]$, $[125, 250]$. - Problem 11: $f(-2) = -10$, $g(-2) = 35$. - Problem 12: $f(\frac{3}{2}) = 177$, range $(-399, 1905)$, $a = 9.2$.