1. **Plot the graphs of the functions:** a) The function is $y=3x^2-7$. - This is a quadratic function with a parabola opening upwards. b) The function is $y=x^3-4x^2+4x$. - This is a cubic function with possible turning points. 2. **Calculate the inverse of the functions:** a) Given $f(x)=\frac{4x+2}{5}$. - To find the inverse, set $y=\frac{4x+2}{5}$. - Multiply both sides by 5: $5y=4x+2$. - Solve for $x$: $4x=5y-2$. - $x=\frac{5y-2}{4}$. - Swap $x$ and $y$ to get the inverse function: $$f^{-1}(x)=\frac{5x-2}{4}$$ b) Given $f(x)=(x+3)^3-5$. - Set $y=(x+3)^3-5$. - Add 5 to both sides: $y+5=(x+3)^3$. - Take cube root: $\sqrt[3]{y+5}=x+3$. - Solve for $x$: $x=\sqrt[3]{y+5}-3$. - Swap $x$ and $y$: $$f^{-1}(x)=\sqrt[3]{x+5}-3$$ 3. **Evaluate the limits:** a) $\lim_{x\to -5} \frac{x^2-25}{x^2+2x-15}$ - Factor numerator: $x^2-25=(x-5)(x+5)$. - Factor denominator: $x^2+2x-15=(x+5)(x-3)$. - Simplify: $\frac{(x-5)(x+5)}{(x+5)(x-3)}=\frac{x-5}{x-3}$ for $x\neq -5$. - Substitute $x=-5$: $\frac{-5-5}{-5-3}=\frac{-10}{-8}=\frac{5}{4}$. b) $\lim_{x\to 0} \frac{(6+x)^2-36}{x}$ - Expand numerator: $(6+x)^2=36+12x+x^2$. - Numerator becomes $36+12x+x^2-36=12x+x^2$. - Expression: $\frac{12x+x^2}{x}=12+x$ for $x\neq 0$. - Substitute $x=0$: $12+0=12$. c) $\lim_{x\to 5} (10 + |x-5|)$ - As $x\to 5$, $|x-5|\to 0$. - Limit is $10+0=10$. d) $\lim_{x\to 1} f(x)$ where $$f(x)=\begin{cases} 7-4x, & x<1 \\ x^2+2, & x\geq 1 \end{cases}$$ - Left-hand limit: $\lim_{x\to 1^-} 7-4x=7-4(1)=3$. - Right-hand limit: $\lim_{x\to 1^+} x^2+2=1+2=3$. - Since both limits equal 3, the limit exists and is 3. **Final answers:** - Inverse functions: $$f^{-1}(x)=\frac{5x-2}{4}$$ and $$f^{-1}(x)=\sqrt[3]{x+5}-3$$ - Limits: $$\lim_{x\to -5} \frac{x^2-25}{x^2+2x-15}=\frac{5}{4}$$ $$\lim_{x\to 0} \frac{(6+x)^2-36}{x}=12$$ $$\lim_{x\to 5} (10 + |x-5|)=10$$ $$\lim_{x\to 1} f(x)=3$$