Curve sketching is a fundamental technique in calculus used to visualize the behavior of a function. By analyzing key features such as intercepts, asymptotes, critical points, and concavity, we can accurately sketch the graph of a function without relying solely on plotting individual points.
The development of curve sketching techniques is closely tied to the history of calculus itself. In the 17th century, mathematicians like Isaac Newton and Gottfried Wilhelm Leibniz developed the fundamental principles of calculus, which provided the tools necessary to analyze and understand the behavior of curves. Early applications focused on geometric problems and the study of motion.
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โ Problem 1
Sketch the curve of the function $f(x) = x^3 - 6x^2 + 9x$.
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Solution
1. Domain: All real numbers.
2. Intercepts: $x(x^2 - 6x + 9) = x(x-3)^2$. $x$-intercepts are $x = 0$ and $x = 3$. $y$-intercept is $f(0) = 0$.
3. Asymptotes: None.
4. First Derivative: $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$. Critical points: $x = 1, 3$.
5. Increasing/Decreasing: $f'(x) > 0$ for $x < 1$ and $x > 3$ (increasing). $f'(x) < 0$ for $1 < x < 3$ (decreasing).
6. Second Derivative: $f''(x) = 6x - 12 = 6(x-2)$. Inflection point: $x = 2$.
7. Concavity: $f''(x) < 0$ for $x < 2$ (concave down). $f''(x) > 0$ for $x > 2$ (concave up).
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โ Problem 2
Sketch the curve of the function $f(x) = \frac{x}{x^2 + 1}$.
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Solution
1. Domain: All real numbers.
2. Intercepts: $x$-intercept is $x = 0$. $y$-intercept is $f(0) = 0$.
3. Asymptotes: Horizontal asymptote at $y = 0$ as $x \rightarrow \pm \infty$.
4. First Derivative: $f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}$. Critical points: $x = -1, 1$.
5. Increasing/Decreasing: $f'(x) > 0$ for $-1 < x < 1$ (increasing). $f'(x) < 0$ for $x < -1$ and $x > 1$ (decreasing).
6. Second Derivative: $f''(x) = \frac{2x(x^2 - 3)}{(x^2 + 1)^3}$. Inflection points: $x = 0, \pm \sqrt{3}$.
7. Concavity: Analyze the sign of $f''(x)$ in the intervals determined by the inflection points.
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โ Problem 3
Sketch the curve of the function $f(x) = x e^{-x}$.
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Solution
1. Domain: All real numbers.
2. Intercepts: $x$-intercept is $x = 0$. $y$-intercept is $f(0) = 0$.
3. Asymptotes: Horizontal asymptote at $y = 0$ as $x \rightarrow \infty$.
4. First Derivative: $f'(x) = e^{-x} - x e^{-x} = e^{-x}(1 - x)$. Critical point: $x = 1$.
5. Increasing/Decreasing: $f'(x) > 0$ for $x < 1$ (increasing). $f'(x) < 0$ for $x > 1$ (decreasing).
6. Second Derivative: $f''(x) = x e^{-x} - 2 e^{-x} = e^{-x}(x - 2)$. Inflection point: $x = 2$.
7. Concavity: $f''(x) < 0$ for $x < 2$ (concave down). $f''(x) > 0$ for $x > 2$ (concave up).
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โ Problem 4
Sketch the curve of the function $f(x) = \frac{1}{x^2 - 1}$.
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Solution
1. Domain: $x \neq \pm 1$.
2. Intercepts: $y$-intercept is $f(0) = -1$. No $x$-intercepts.
3. Asymptotes: Vertical asymptotes at $x = \pm 1$. Horizontal asymptote at $y = 0$.
4. First Derivative: $f'(x) = \frac{-2x}{(x^2 - 1)^2}$. Critical point: $x = 0$.
5. Increasing/Decreasing: $f'(x) > 0$ for $x < -1$ and $-1 < x < 0$ (increasing). $f'(x) < 0$ for $0 < x < 1$ and $x > 1$ (decreasing).
6. Second Derivative: $f''(x) = \frac{6x^2 + 2}{(x^2 - 1)^3}$. No inflection points.
7. Concavity: Analyze the sign of $f''(x)$ in the intervals determined by the vertical asymptotes.
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โ Problem 5
Sketch the curve of the function $f(x) = x^4 - 2x^2$.
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Solution
1. Domain: All real numbers.
2. Intercepts: $x^2(x^2 - 2) = 0$. $x$-intercepts are $x = 0, \pm \sqrt{2}$. $y$-intercept is $f(0) = 0$.
3. Asymptotes: None.
4. First Derivative: $f'(x) = 4x^3 - 4x = 4x(x^2 - 1)$. Critical points: $x = 0, \pm 1$.
5. Increasing/Decreasing: Analyze the sign of $f'(x)$ in the intervals determined by the critical points.
6. Second Derivative: $f''(x) = 12x^2 - 4 = 4(3x^2 - 1)$. Inflection points: $x = \pm \frac{1}{\sqrt{3}}$.
7. Concavity: Analyze the sign of $f''(x)$ in the intervals determined by the inflection points.
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โ Problem 6
Sketch the curve of the function $f(x) = \ln(x^2 + 1)$.
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Solution
1. Domain: All real numbers.
2. Intercepts: $x$-intercept is $x = 0$. $y$-intercept is $f(0) = 0$.
3. Asymptotes: None.
4. First Derivative: $f'(x) = \frac{2x}{x^2 + 1}$. Critical point: $x = 0$.
5. Increasing/Decreasing: $f'(x) < 0$ for $x < 0$ (decreasing). $f'(x) > 0$ for $x > 0$ (increasing).
6. Second Derivative: $f''(x) = \frac{2 - 2x^2}{(x^2 + 1)^2}$. Inflection points: $x = \pm 1$.
7. Concavity: Analyze the sign of $f''(x)$ in the intervals determined by the inflection points.
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โ Problem 7
Sketch the curve of the function $f(x) = \frac{x^2}{x^2 - 4}$.
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Solution
1. Domain: $x \neq \pm 2$.
2. Intercepts: $x$-intercept is $x = 0$. $y$-intercept is $f(0) = 0$.
3. Asymptotes: Vertical asymptotes at $x = \pm 2$. Horizontal asymptote at $y = 1$.
4. First Derivative: $f'(x) = \frac{-8x}{(x^2 - 4)^2}$. Critical point: $x = 0$.
5. Increasing/Decreasing: $f'(x) > 0$ for $x < -2$ and $-2 < x < 0$ (increasing). $f'(x) < 0$ for $0 < x < 2$ and $x > 2$ (decreasing).
6. Second Derivative: $f''(x) = \frac{8(3x^2 + 4)}{(x^2 - 4)^3}$. No inflection points.
7. Concavity: Analyze the sign of $f''(x)$ in the intervals determined by the vertical asymptotes.
Mastering curve sketching requires a solid understanding of calculus concepts and a systematic approach. By following the steps outlined above and practicing with various examples, you can develop the skills necessary to accurately sketch the graphs of functions and gain valuable insights into their behavior. Keep practicing, and you'll become a curve sketching pro in no time!