1. Problem 1: Given $f(x) = 5x - a$, $g(x) = x^2$, and $fg(4) = 5$, find $a$. 2. Recall that $fg(4) = f(g(4)) = f(16) = 5$. 3. Substitute $x=16$ into $f(x)$: $f(16) = 5(16) - a = 80 - a$. 4. Set equal to 5: $$80 - a = 5$$ 5. Solve for $a$: $$a = 80 - 5 = 75$$. 6. Problem 2: Given $f(x) = \sqrt{\frac{x - 7}{9}}$, find $|x|$ if $f^{-1}(x) = 583$. 7. To find $f^{-1}(x)$, swap $x$ and $y$ in $y = \sqrt{\frac{x - 7}{9}}$: $$x = \sqrt{\frac{y - 7}{9}}$$ 8. Square both sides: $$x^2 = \frac{y - 7}{9}$$ 9. Multiply both sides by 9: $$9x^2 = y - 7$$ 10. Solve for $y$: $$y = 9x^2 + 7$$ 11. So, $f^{-1}(x) = 9x^2 + 7$. 12. Given $f^{-1}(x) = 583$, set: $$9x^2 + 7 = 583$$ 13. Subtract 7: $$9x^2 = 576$$ 14. Divide both sides by 9: $$\cancel{9}x^2 = \frac{576}{\cancel{9}}$$ $$x^2 = 64$$ 15. Take square root: $$x = \pm 8$$ 16. Find $|x|$: $$|x| = 8$$ 17. Problem 3: Given $f(x) = \frac{x}{x + c}$, find $c$ if $f(x) \equiv f^{-1}(x)$. 18. To find $f^{-1}(x)$, set $y = \frac{x}{x + c}$ and solve for $x$: $$y = \frac{x}{x + c}$$ 19. Multiply both sides by $x + c$: $$y(x + c) = x$$ 20. Distribute: $$yx + yc = x$$ 21. Rearrange terms: $$yx - x = -yc$$ 22. Factor $x$: $$x(y - 1) = -yc$$ 23. Solve for $x$: $$x = \frac{-yc}{y - 1}$$ 24. Swap $x$ and $y$ to get $f^{-1}(x)$: $$f^{-1}(x) = \frac{-xc}{x - 1}$$ 25. Since $f(x) \equiv f^{-1}(x)$, set: $$\frac{x}{x + c} = \frac{-xc}{x - 1}$$ 26. Cross multiply: $$(x)(x - 1) = (-xc)(x + c)$$ 27. Expand both sides: $$x^2 - x = -x^2 c - x c^2$$ 28. Bring all terms to one side: $$x^2 - x + x^2 c + x c^2 = 0$$ 29. Factor $x$: $$x^2 + x^2 c + x c^2 - x = 0$$ 30. Group terms: $$x^2(1 + c) + x(c^2 - 1) = 0$$ 31. For this to hold for all $x$, coefficients must be zero: $$1 + c = 0 \implies c = -1$$ 32. Check second coefficient: $$c^2 - 1 = (-1)^2 - 1 = 1 - 1 = 0$$ 33. Both conditions satisfied with $c = -1$. Final answers: - $a = 75$ - $|x| = 8$ - $c = -1$