1. **Problem Statement:** We need to sketch a graph of a function $f(x)$ with the following properties: - $f(x)$ is defined for all real $x$ except $x = -2$. - The only root of $f(x)$ is at $x=0$. - Points on the graph include $(-4,1)$, $(2,2)$, and $(3,3)$. - $f$ is continuous on $(-\infty,-2)$, $(-2,2)$, and $[2,+\infty)$. - $f$ is differentiable everywhere except at $x=-2, 2, 3$. - $f'(x) < 0$ on $(-\infty,-4)$ and $(3,+\infty)$; $f'(x) > 0$ on $(-4,-2)$, $(-2,2)$, and $(2,3)$. - $f''(x) < 0$ on $(-\infty,-5)$, $(-\infty,2,0)$, and $(2,3)$; $f''(x) > 0$ on $(-5,-2)$, $(0,2)$, and $(2,+\infty)$. - The only zero of $f'(x)$ is at $x=-4$. - Limits: $\lim_{x\to -\infty} f(x) = +\infty$, $\lim_{x\to 4^-} f(x) = -\infty$, $\lim_{x\to 2^-} f'(x) = +\infty$, $\lim_{x\to +\infty} f(x) = 2$, and $\lim_{x\to +\infty} f'(x) = 1$. 2. **Understanding the problem:** - The function has a vertical discontinuity at $x=-2$. - It crosses the $x$-axis only at $0$. - It passes through given points. - The derivative's sign tells us where the function is increasing or decreasing. - The second derivative's sign tells us about concavity. - Limits describe end behavior and behavior near discontinuities. 3. **Key formulas and rules:** - Increasing if $f'(x) > 0$, decreasing if $f'(x) < 0$. - Concave up if $f''(x) > 0$, concave down if $f''(x) < 0$. - Continuity means no breaks except at $x=-2$. - Differentiability fails at $x=-2, 2, 3$. 4. **Step-by-step analysis:** - **Domain and discontinuity:** $f$ is undefined at $x=-2$, so expect a vertical asymptote or jump there. - **Roots:** $f(0) = 0$ and no other zeros. - **Points:** $f(-4) = 1$, $f(2) = 2$, $f(3) = 3$. - **Continuity:** Continuous on $(-\infty,-2)$, $(-2,2)$, and $[2,+\infty)$. - **Differentiability:** Not differentiable at $x=-2, 2, 3$. - **Derivative sign:** - $f'(x) < 0$ on $(-\infty,-4)$ and $(3,+\infty)$ means $f$ is decreasing there. - $f'(x) > 0$ on $(-4,-2)$, $(-2,2)$, and $(2,3)$ means $f$ is increasing there. - **Second derivative sign:** - $f''(x) < 0$ on $(-\infty,-5)$, $(-\infty,2,0)$, and $(2,3)$ means concave down there. - $f''(x) > 0$ on $(-5,-2)$, $(0,2)$, and $(2,+\infty)$ means concave up there. - **Critical points:** The only zero of $f'(x)$ is at $x=-4$, so $f$ has a local max or min there. - **Limits:** - $\lim_{x\to -\infty} f(x) = +\infty$ means the graph goes up to infinity on the far left. - $\lim_{x\to 4^-} f(x) = -\infty$ means the graph goes down to negative infinity approaching $4$ from the left. - $\lim_{x\to 2^-} f'(x) = +\infty$ means the slope becomes very steep positive near $2$ from the left. - $\lim_{x\to +\infty} f(x) = 2$ means the graph approaches $y=2$ as $x$ goes to infinity. - $\lim_{x\to +\infty} f'(x) = 1$ means the slope approaches $1$ as $x$ goes to infinity. 5. **Sketching the graph:** - Start from far left: $f(x) \to +\infty$ and $f'(x) < 0$ on $(-\infty,-4)$ so the graph decreases from $+\infty$ to $f(-4)=1$. - At $x=-4$, $f'(x)=0$ (local extremum), and since $f'(x)$ changes from negative to positive, this is a local minimum. - On $(-4,-2)$, $f'(x) > 0$ so $f$ increases from $1$ to near the discontinuity at $x=-2$. - At $x=-2$, $f$ is not defined, so a vertical asymptote or jump. - On $(-2,2)$, $f'(x) > 0$ and $f$ is continuous, so $f$ increases from near $-\infty$ (due to discontinuity) to $f(2)=2$. - At $x=2$, $f$ is continuous but not differentiable, slope tends to $+\infty$ from left. - On $(2,3)$, $f'(x) > 0$ so $f$ increases from $2$ to $3$. - At $x=3$, $f$ is not differentiable. - On $(3,+\infty)$, $f'(x) < 0$ so $f$ decreases from $3$ towards $2$ as $x \to +\infty$. - The concavity changes as per $f''(x)$ intervals. 6. **Summary:** - The graph has a vertical asymptote at $x=-2$. - It has a local minimum at $x=-4$. - It crosses the $x$-axis only at $0$. - It passes through $(2,2)$ and $(3,3)$. - It increases and decreases according to the derivative signs. - It approaches $y=2$ as $x \to +\infty$. This detailed analysis guides the sketch of the graph.