Algebra Review 9C22F7

1. **Find the inverse of** $f(x) = \frac{1}{3}x - 4$. Start by replacing $f(x)$ with $y$: $$y = \frac{1}{3}x - 4$$ To find the inverse, swap $x$ and $y$: $$x = \frac{1}{3}y - 4$$ Add 4 to both sides: $$x + 4 = \frac{1}{3}y$$ Multiply both sides by 3 to isolate $y$: $$3(x + 4) = y$$ Simplify: $$y = 3x + 12$$ So, the inverse function is: $$f^{-1}(x) = 3x + 12$$ 2. **Graph the original function** $f(x) = \frac{1}{3}x - 4$ **and its inverse** $f^{-1}(x) = 3x + 12$. - The graph of $f(x)$ is a line with slope $\frac{1}{3}$ and y-intercept $-4$. - The graph of $f^{-1}(x)$ is a line with slope $3$ and y-intercept $12$. - They are reflections of each other across the line $y = x$. 3. **Find the inverse of** $y = -\frac{2}{5}x + 4$. Swap $x$ and $y$: $$x = -\frac{2}{5}y + 4$$ Subtract 4 from both sides: $$x - 4 = -\frac{2}{5}y$$ Multiply both sides by $-\frac{5}{2}$ to isolate $y$: $$y = -\frac{5}{2}(x - 4)$$ Simplify: $$y = -\frac{5}{2}x + 10$$ So, the inverse function is: $$f^{-1}(x) = -\frac{5}{2}x + 10$$ 4. **Graph the original function and inverse from #3 and check if they are inverses using composition.** - Original: $f(x) = -\frac{2}{5}x + 4$ - Inverse: $f^{-1}(x) = -\frac{5}{2}x + 10$ Check composition $f(f^{-1}(x))$: $$f(f^{-1}(x)) = -\frac{2}{5} \left(-\frac{5}{2}x + 10\right) + 4$$ $$= -\frac{2}{5} \times -\frac{5}{2}x - \frac{2}{5} \times 10 + 4$$ $$= (1)x - 4 + 4 = x$$ Check composition $f^{-1}(f(x))$: $$f^{-1}(f(x)) = -\frac{5}{2} \left(-\frac{2}{5}x + 4\right) + 10$$ $$= -\frac{5}{2} \times -\frac{2}{5}x - \frac{5}{2} \times 4 + 10$$ $$= (1)x - 10 + 10 = x$$ Since both compositions equal $x$, the functions are inverses. 5. **Determine if** $f(x) = \frac{1}{2}x - \frac{3}{2}$ **and** $g(x) = 2x + 3$ **are inverses.** Compute $f(g(x))$: $$f(g(x)) = \frac{1}{2}(2x + 3) - \frac{3}{2} = x + \frac{3}{2} - \frac{3}{2} = x$$ Compute $g(f(x))$: $$g(f(x)) = 2\left(\frac{1}{2}x - \frac{3}{2}\right) + 3 = x - 3 + 3 = x$$ Both compositions equal $x$, so **A) Yes**, they are inverses. 6. **Evaluate the piecewise function:** $$f(x) = \begin{cases} 5x - 1, & x < -5 \\ -2x + 3, & x \geq -5 \end{cases}$$ a. $f(4)$: Since $4 \geq -5$, use $-2x + 3$: $$f(4) = -2(4) + 3 = -8 + 3 = -5$$ b. $f(-5)$: Since $-5 \geq -5$, use $-2x + 3$: $$f(-5) = -2(-5) + 3 = 10 + 3 = 13$$ c. $f(-1)$: Since $-1 \geq -5$, use $-2x + 3$: $$f(-1) = -2(-1) + 3 = 2 + 3 = 5$$ 7. **Find** $(g \circ f)(x)$ **where** $f(x) = -2x - 3$ **and** $g(x) = 3x + 1$. Compute $g(f(x))$: $$g(f(x)) = 3(-2x - 3) + 1 = -6x - 9 + 1 = -6x - 8$$ 8. **Find** $f(g(-2))$ **where** $f(x) = 3x - 6$ **and** $g(x) = -5x + 7$. First find $g(-2)$: $$g(-2) = -5(-2) + 7 = 10 + 7 = 17$$ Then find $f(17)$: $$f(17) = 3(17) - 6 = 51 - 6 = 45$$ 9. **Sketch the piecewise function:** $$f(x) = \begin{cases} -\frac{1}{3}x - 4, & x \leq 3 \\ 3, & x > 3 \end{cases}$$ - For $x \leq 3$, graph the line $y = -\frac{1}{3}x - 4$ with a closed point at $(3, -5)$. - For $x > 3$, graph the horizontal line $y = 3$ with an open point at $(3, 3)$. 10. **Given graph of piecewise function** $g(x)$: $$g(x) = \begin{cases} A, & x \leq 1 \\ B, & x > 1 \end{cases}$$ From the description, identify equations for $A$ and $B$ based on the graph segments. 11. **Given graph of piecewise function** $g(x)$: $$g(x) = \begin{cases} A, & -8 \leq x \leq 2 \\ B, & -2 \leq x \leq 4 \\ C, & 4 < x \leq 10 \end{cases}$$ Identify equations for $A$, $B$, and $C$ from the graph segments. 12. **Analyze how total hours spent studying affect final exam grade.** Data points: $$(2,5), (5,7), (1,4), (0,3), (4,6), (7,8.5), (3,5.5)$$ a. Using linear regression, the line of best fit is: $$y = 0.75x + 3$$ b. Estimate exam score for 8 hours: $$y = 0.75(8) + 3 = 6 + 3 = 9$$ c. Correlation coefficient $r \approx 0.98$ indicates a very strong positive linear relationship between hours studied and exam score.