1. Problem: Given $f(x)=-7x^2-5$ and $g(x)=2x-1$, solve $f(g(x))=-68$. 1. Formula and rule: $f(g(x))$ means substitute $g(x)$ into $f$, so compute $f(g(x))$. 1. Work: $g(x)=2x-1$ so $f(g(x))=-7(2x-1)^2-5$. 1. Work: $(2x-1)^2=4x^2-4x+1$ so $f(g(x))=-7(4x^2-4x+1)-5=-28x^2+28x-12$. 1. Solve: $-28x^2+28x-12=-68$ gives $-28x^2+28x+56=0$. 1. Simplify: divide by $-28$ to get $x^2-x-2=0$. 1. Factor: $(x-2)(x+1)=0$ so $x=2$ or $x=-1$. 2. Problem: Solve $|2x-3|\le 9$. 2. Rule: For $|A|\le B$ with $B\ge 0$ we have $-B\le A\le B$. 2. Work: $-9\le 2x-3\le 9$. 2. Solve: add $3$ to each part to get $-6\le 2x\le 12$. 2. Divide by $2$ to obtain $-3\le x\le 6$. 3. Problem: Show that $x-3$ is a factor of $f(x)=2x^3-9x^2+10x-3$. 3. Rule: $x-a$ is a factor iff $f(a)=0$ (Remainder Theorem). 3. Work: compute $f(3)=2\cdot3^3-9\cdot3^2+10\cdot3-3=54-81+30-3=0$. 3. Conclusion: $f(3)=0$ so $x-3$ is a factor. 4. Problem: Hence solve $f(x)=0$ for $f(x)=2x^3-9x^2+10x-3$. 4. Work: divide by $x-3$ to get quotient $2x^2-3x+1$. 4. Factor quadratic: $2x^2-3x+1=(2x-1)(x-1)$. 4. Roots: $x=3,\;x=\frac{1}{2},\;x=1$. 5. Problem: The midpoint of $PQ$ is $(-7,-2)$ and $P=(6,8)$; find $Q$. 5. Formula: midpoint $M=\left(\frac{x_P+x_Q}{2},\frac{y_P+y_Q}{2}\right)$. 5. Work: set $\frac{6+x_Q}{2}=-7$ and $\frac{8+y_Q}{2}=-2$. 5. Solve: $6+x_Q=-14$ so $x_Q=-20$; $8+y_Q=-4$ so $y_Q=-12$. 5. Answer: $Q=(-20,-12)$. 6. Problem: Solve $5^{-(5x+8)}=25^{2x+5}$. 6. Rule: $25=5^2$ so equate powers of 5 when bases are positive and equal. 6. Work: $5^{-(5x+8)}=(5^2)^{2x+5}=5^{4x+10}$. 6. Equate exponents: $-(5x+8)=4x+10$ so $-5x-8=4x+10$. 6. Solve: $-9x=18$ so $x=-2$. 7. Problem: Solve $4\log_9(x-2)=\log_9 3+5$ with domain considerations. 7. Rule: $a\log_b(c)=\log_b(c^a)$ and $\log_b(u)+\log_b(v)=\log_b(uv)$. 7. Work: left gives $\log_9((x-2)^4)$ and right write $5=\log_9(9^5)$ so right is $\log_9(3\cdot9^5)$. 7. Simplify numeric: $9^5=59049$ and $3\cdot9^5=177147=3^{11}$ so equation is $(x-2)^4=3^{11}$. 7. Solve principal real root: $x-2=3^{11/4}$ or $x-2=-3^{11/4}$ but domain of $\log_9(x-2)$ requires $x-2>0$. 7. Conclusion: only $x=2+3^{11/4}$ is valid. 8. Problem: Sketch $y=e^x$ and its inverse on same axes and show intercepts. 8. Facts: $y=e^x$ has domain $(-\infty,\infty)$ and range $(0,\infty)$ and intercept $(0,1)$. 8. Facts: inverse is $y=\ln x$ with domain $(0,\infty)$ and range $(-\infty,\infty)$ and intercept $(1,0)$. 8. Rule: a function and its inverse are symmetric about the line $y=x$. 8. Description: sketch $y=e^x$ passing through $(0,1)$ growing to the right and approaching $y=0$ to the left, and sketch $y=\ln x$ reflected across $y=x$ passing through $(1,0)$. 9. Problem: Using Cramer or otherwise, determine $y$ for the system $5x-3y+5z=-34$, $-x+y+z=4$, $2x-3y-3z=-13$. 9. Method: use elimination to find $y$ directly. 9. Work: from $-x+y+z=4$ get $x=y+z-4$. 9. Substitute into first and third to get $y+5z=-7$ and $y+z=5$. 9. Subtract to get $4z=-12$ so $z=-3$ and then $y+(-3)=5$ so $y=8$. 9. Answer: $y=8$. 10. Problem: Define injective and state whether $f(x)=2-x$ is injective. 10. Definition: a function is injective (one-to-one) if distinct inputs give distinct outputs. 10. Reason: $f(x)=2-x$ is linear with slope $-1\neq 0$ so it is strictly monotonic and therefore injective. 11. Problem: Given $f(x)=2-x$ and $g(x)=2^x$, solve $g(f(x))=16$. 11. Work: $g(f(x))=2^{f(x)}=2^{2-x}$ and $16=2^4$. 11. Equate exponents: $2-x=4$ so $x=-2$. 12. Problem: Determine the range of real $x$ with $3x^2+8x\le 3$. 12. Work: bring to one side $3x^2+8x-3\le 0$. 12. Discriminant: $\Delta=8^2-4\cdot3\cdot(-3)=64+36=100$ so roots are $\frac{-8\pm10}{6}$ giving $x=-3$ and $x=\frac{1}{3}$. 12. Conclusion: quadratic opens up so $-3\le x\le \frac{1}{3}$. 13. Problem: A board of length 258 cm is cut into 12 pieces in arithmetic progression and the sum of the first 3 pieces is 24 cm; find the first piece. 13. Let first term be $a$ and common difference $d$. 13. Sum of first 3: $a+(a+d)+(a+2d)=3a+3d=24$ so $a+d=8$. 13. Sum of 12 terms: $\frac{12}{2}(2a+11d)=6(2a+11d)=258$ so $2a+11d=43$. 13. Solve with $a=8-d$ to get $2(8-d)+11d=43$ so $9d=27$ giving $d=3$ and then $a=5$. 13. Answer: first piece is $5$ cm. 14. Problem: Let $f(x)=-x^3+ax^2-5x-12$. If $x+1$ is a factor show $a=6$. 14. Rule: substitute root $x=-1$ into $f$ and set equal to $0$. 14. Work: $f(-1)=-(-1)^3+a(-1)^2-5(-1)-12=1+a+5-12=a-6$. 14. Set $a-6=0$ so $a=6$. 15. Problem: Hence determine the other two factors when $a=6$. 15. Work: with $a=6$ we have $f(x)=-x^3+6x^2-5x-12$ and dividing by $x+1$ gives quotient $-x^2+7x-12$. 15. Factor: $-x^2+7x-12=-(x^2-7x+12)=-(x-3)(x-4)$. 15. Thus the other linear factors (up to sign) are $x-3$ and $x-4$. 16. Problem: Solve $\sqrt{2}\cos(\theta+\pi/3)=1$ for $-\pi<\theta<\pi$. 16. Work: divide to get $\cos(\theta+\pi/3)=\frac{1}{\sqrt{2}}=\cos\frac{\pi}{4}$. 16. General solutions: $\theta+\pi/3=\pm\frac{\pi}{4}+2k\pi$. 16. Solve for principal values with $k=0$: $\theta=-\frac{\pi}{3}+\frac{\pi}{4}=-\frac{\pi}{12}$ and $\theta=-\frac{\pi}{3}-\frac{\pi}{4}=-\frac{7\pi}{12}$. 16. Both lie in $(-\pi,\pi)$ so solutions are $\theta=-\frac{\pi}{12},-\frac{7\pi}{12}$.