1. **Problem Statement:** We have the function $g(x) = 3x^2 + 2$ and a set of points for $f(x)$ given as: $$\begin{array}{c|c} x & f(x) \\\hline -2 & 2 \\ -1 & 1 \\ 2 & 3 \\ 3 & 2 \end{array}$$ We need to find $g(3)$, $g(-2)$, $g(1)$, and $g(0)$, determine the domain and inverse of $g$, and analyze the function $f$ on given intervals. 2. **Evaluate $g(x)$ at given points:** - $g(3) = 3(3)^2 + 2 = 3 \times 9 + 2 = 27 + 2 = 29$ - $g(-2) = 3(-2)^2 + 2 = 3 \times 4 + 2 = 12 + 2 = 14$ - $g(1) = 3(1)^2 + 2 = 3 \times 1 + 2 = 3 + 2 = 5$ - $g(0) = 3(0)^2 + 2 = 0 + 2 = 2$ 3. **Domain and inverse of $g(x)$:** - The domain of $g$ is all real numbers $\mathbb{R}$ because it is a polynomial. - To find the inverse, solve $y = 3x^2 + 2$ for $x$: $$y - 2 = 3x^2 \implies x^2 = \frac{y - 2}{3} \implies x = \pm \sqrt{\frac{y - 2}{3}}$$ - Since $g$ is not one-to-one over all $\mathbb{R}$, restrict domain to $[2,3]$ or $[-1,2]$ to have an inverse. 4. **Domain and range of $f$ on intervals:** - Given points suggest $f$ is defined at $x = -2, -1, 2, 3$ with corresponding $f(x)$ values. - On $[-1,2]$, $f$ values are $1$ at $-1$ and $3$ at $2$. - The inverse of $f$ on $[2,3]$ can be found by swapping $x$ and $f(x)$ values. 5. **Solve quadratic $P(x) = x^2 + 2x - 3$:** - Set $P(x) = 0$: $$x^2 + 2x - 3 = 0$$ - Calculate discriminant: $$\Delta = b^2 - 4ac = 2^2 - 4 \times 1 \times (-3) = 4 + 12 = 16$$ - Roots: $$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-2 \pm 4}{2}$$ $$x_1 = \frac{-2 + 4}{2} = 1, \quad x_2 = \frac{-2 - 4}{2} = -3$$ 6. **System of equations:** - Given: $$\begin{cases} 3x - y = 5 \\ x + 2y = 4 \end{cases}$$ - Solve for $x$ and $y$: From first: $y = 3x - 5$ Substitute into second: $$x + 2(3x - 5) = 4 \implies x + 6x - 10 = 4 \implies 7x = 14 \implies x = 2$$ Then $y = 3(2) - 5 = 6 - 5 = 1$ **Final answers:** - $g(3) = 29$, $g(-2) = 14$, $g(1) = 5$, $g(0) = 2$ - Domain of $g$ is $\mathbb{R}$; inverse exists on restricted domains. - Roots of $P(x)$ are $1$ and $-3$. - Solution to system: $(x,y) = (2,1)$.