Topics Covered:
Checking for symmetry can simplify the graphing process by reducing the amount of analysis needed.
Even Function: A function \(f(x)\) is even if:
\[f(-x) = f(x) \text{ for all } x\]Even functions have y-axis symmetry. The graph is symmetric about the y-axis, meaning if you fold it along the y-axis, both halves match perfectly.
Examples of even functions:
Graph Description: The function \(y = x^2\) is a classic example of an even function. The parabola is symmetric about the y-axis, with the left side being a mirror image of the right side.
Odd Function: A function \(f(x)\) is odd if:
\[f(-x) = -f(x) \text{ for all } x\]Odd functions have origin symmetry. The graph is symmetric about the origin, meaning if you rotate it 180° about the origin, it looks the same.
Examples of odd functions:
Graph Description: The function \(y = x^3\) is a classic example of an odd function. If you rotate the graph 180° around the origin, it appears unchanged. The portion in the third quadrant is a rotated version of the portion in the first quadrant.
Step 1: Domain
Domain: \((-\infty, \infty)\) (all real numbers)
Step 2: Symmetry
Check if even:
\[f(-x) = e^{-(-x)^2} = e^{-x^2} = f(x)\]Since \(f(-x) = f(x)\), this is an even function with y-axis symmetry.
Step 3: Intercepts
x-intercept: Solve \(f(x) = e^{-x^2} = 0\)
Since \(e^{-x^2} > 0\) for all \(x\), there are no x-intercepts.
y-intercept:
\[f(0) = e^{-0^2} = e^0 = 1\]Step 4: Vertical Asymptotes
No vertical asymptotes (exponential functions have no vertical asymptotes)
Step 5: End Behavior
\[\lim_{x \to \infty} e^{-x^2} = \lim_{x \to \infty} \frac{1}{e^{x^2}} = 0\]Therefore, \(y = 0\) is a horizontal asymptote.
Step 6: First Derivative
Using the chain rule:
\[f'(x) = \frac{d}{dx}[e^{-x^2}] = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}\]Critical points: Set \(f'(x) = 0\)
\[-2xe^{-x^2} = 0\]Since \(e^{-x^2} \neq 0\), we have \(-2x = 0\), so \(x = 0\) is the only critical point.
Sign analysis of \(f'(x)\):
| Interval | Sign of \(-2x\) | Sign of \(f'(x)\) | Behavior |
|---|---|---|---|
| \((-\infty, 0)\) | \(+\) | \(+\) | Increasing ↑ |
| \((0, \infty)\) | \(-\) | \(-\) | Decreasing ↓ |
Conclusion: \(x = 0\) is a local maximum.
Step 7: Second Derivative
Using the product rule on \(f'(x) = -2xe^{-x^2}\):
\[f''(x) = -2e^{-x^2} + (-2x)(-2x)e^{-x^2}\] \[= -2e^{-x^2} + 4x^2e^{-x^2}\] \[= e^{-x^2}(4x^2 - 2)\] \[= 2e^{-x^2}(2x^2 - 1)\]Inflection points: Set \(f''(x) = 0\)
\[2e^{-x^2}(2x^2 - 1) = 0\]Since \(e^{-x^2} \neq 0\), we have:
\[2x^2 - 1 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2}\]Sign analysis of \(f''(x)\):
| Interval | Sign of \((2x^2-1)\) | Sign of \(f''(x)\) | Concavity |
|---|---|---|---|
| \((-\infty, -\frac{\sqrt{2}}{2})\) | \(+\) | \(+\) | Concave Up ∪ |
| \((-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) | \(-\) | \(-\) | Concave Down ∩ |
| \((\frac{\sqrt{2}}{2}, \infty)\) | \(+\) | \(+\) | Concave Up ∪ |
Conclusion: Inflection points at \(x = -\frac{\sqrt{2}}{2}\) and \(x = \frac{\sqrt{2}}{2}\).
Complete Sketch Description: The graph of \(f(x) = e^{-x^2}\) is a bell-shaped curve centered at the origin. It has a maximum at \((0, 1)\), approaches the x-axis asymptotically as \(x \to \pm\infty\), and has inflection points at approximately \((\pm 0.707, 0.606)\). The curve is symmetric about the y-axis, concave down in the middle region and concave up on the outer regions.
Step 1: Symmetry
Check if odd:
\[f(-x) = \sin(-x) - (-x) = -\sin(x) + x = -(\sin(x) - x) = -f(x)\]Since \(f(-x) = -f(x)\), this is an odd function with origin symmetry.
Step 2: Intercepts
x-intercept: Solve \(\sin(x) - x = 0\), which means \(\sin(x) = x\)
This transcendental equation has only one solution: \(x = 0\)
y-intercept:
\[f(0) = \sin(0) - 0 = 0\]So \(x = 0\) is the only x-intercept and y-intercept.
Step 3: Vertical Asymptotes
No vertical asymptotes
Step 4: End Behavior
Since our domain is restricted to \([-2\pi, 2\pi]\), we don't need to check behavior as \(x \to \pm\infty\).
Step 5: First Derivative
\[f'(x) = \frac{d}{dx}(\sin(x) - x) = \cos(x) - 1\]Critical points: Set \(f'(x) = 0\)
\[\cos(x) - 1 = 0 \implies \cos(x) = 1\]On \([-2\pi, 2\pi]\): \(x = 0, \pm 2\pi\)
Sign analysis:
Note that \(\cos(x) \leq 1\) for all \(x\), so \(\cos(x) - 1 \leq 0\) always.
Therefore, \(f'(x) < 0\) on \((0, 2\pi)\), meaning the function is decreasing on the entire interval (by symmetry, also decreasing on \((-2\pi, 0)\)).
Step 6: Second Derivative
\[f''(x) = \frac{d}{dx}(\cos(x) - 1) = -\sin(x)\]Inflection points: Set \(f''(x) = 0\)
\[-\sin(x) = 0 \implies x = 0, \pm\pi, \pm 2\pi\]Sign analysis of \(f''(x) = -\sin(x)\):
| Interval | Sign of \(-\sin(x)\) | Concavity |
|---|---|---|
| \((0, \pi)\) | \(-\) | Concave Down ∩ |
| \((\pi, 2\pi)\) | \(+\) | Concave Up ∪ |
Complete Sketch Description: The graph of \(f(x) = \sin(x) - x\) on \([-2\pi, 2\pi]\) passes through the origin and is symmetric about the origin (odd function). The function is continuously decreasing throughout the interval. It has inflection points at \(x = 0, \pm\pi, \pm 2\pi\), alternating between concave down and concave up.
Understanding the problem: We're given the derivative and asked to sketch the original function.
What we know:
Analysis of \(f'(x) = 3x^2\):
Since \(f'(x) = 3x^2 \geq 0\) for all \(x\), and equals 0 only at \(x = 0\), we know:
Analysis of \(f''(x) = 6x\):
What \(f(x)\) looks like:
Since \(f'(x) = 3x^2\), we can find \(f(x)\) by integration:
\[f(x) = \int 3x^2 \, dx = x^3 + C\]where \(C\) is an arbitrary constant.
Sketch Description: The function \(f(x) = x^3 + C\) is a cubic function that increases throughout its domain. It is concave down for \(x < 0\) and concave up for \(x > 0\), with an inflection point at the origin. The exact vertical position depends on the constant \(C\).
Step 1: Find critical points
Set \(f'(x) = 0\):
\[(x+4)(x-1)(x-5) = 0\] \[x = -4, \quad x = 1, \quad x = 5\]Step 2: Determine where \(f(x)\) is increasing/decreasing
Create a sign chart for \(f'(x)\):
| Interval | Sign of each factor | Sign of \(f'(x)\) | Behavior of \(f(x)\) |
|---|---|---|---|
| \((-\infty, -4)\) | \((-)(-)(-) = -\) | \(-\) | Decreasing ↓ |
| \((-4, 1)\) | \((+)(-)(-)= +\) | \(+\) | Increasing ↑ |
| \((1, 5)\) | \((+)(+)(-) = -\) | \(-\) | Decreasing ↓ |
| \((5, \infty)\) | \((+)(+)(+) = +\) | \(+\) | Increasing ↑ |
Step 3: Identify local extrema
Step 4: Analyze concavity using \(f''(x)\)
Since \(f'(x) = (x+4)(x-1)(x-5)\) is a degree 3 polynomial, \(f''(x)\) will be a quadratic (degree 2).
A quadratic has at most 2 roots, meaning \(f''(x) = 0\) at most twice.
Key observation: The function \(f(x)\) is a degree 4 polynomial (since its derivative is degree 3).
Inflection points:
Therefore, there must be at least 2 inflection points, and since \(f''(x)\) is quadratic, there are exactly 2 inflection points.
Complete Sketch Description: The function \(f(x)\) has a wavy shape with three critical points. Starting from the left, the function decreases until \(x = -4\) (local minimum), then increases until \(x = 1\) (local maximum), then decreases until \(x = 5\) (local minimum), and finally increases for \(x > 5\). There are two inflection points between the critical points where the concavity changes.
Key Concepts for Curve Sketching:
1. Symmetry can simplify your work:
2. Always follow the systematic approach:
3. When given \(f'(x)\) instead of \(f(x)\):
4. Connecting derivatives to graph behavior: