Let's solve each part step by step for the function \( f(x) = \frac{x^2}{e^x} \): --- ### **(a) Find all critical values and identify them** **Step 1: Find the derivative** \[ f(x) = \frac{x^2}{e^x} \] Use the quotient rule: \[ f'(x) = \frac{(2x)(e^x) - (x^2)(e^x)}{(e^x)^2} \] \[ = \frac{e^x(2x - x^2)}{e^{2x}} \] \[ = \frac{2x - x^2}{e^x} \] **Step 2: Set \( f'(x) = 0 \) to find critical points** \[ \frac{2x - x^2}{e^x} = 0 \] Since \( e^x > 0 \), set numerator to 0: \[ 2x - x^2 = 0 \] \[ x(2 - x) = 0 \] So, \( x = 0 \) and \( x = 2 \). **Step 3: Classify using the second derivative test** Find \( f''(x) \): \[ f'(x) = \frac{2x - x^2}{e^x} \] Use the quotient rule again: \[ f''(x) = \frac{[2 - 2x]e^x - (2x - x^2)e^x}{e^{2x}} \] \[ = \frac{(2 - 2x - 2x + x^2) e^x}{e^{2x}} \] \[ = \frac{(2 - 4x + x^2) e^x}{e^{2x}} \] \[ = \frac{2 - 4x + x^2}{e^x} \] - At \( x = 0 \): \[ f''(0) = \frac{2}{1} = 2 > 0 \] So, local **minimum** at \( x = 0 \). - At \( x = 2 \): \[ f''(2) = \frac{2 - 8 + 4}{e^2} = \frac{-2}{e^2} < 0 \] So, local **maximum** at \( x = 2 \). --- ### **(b) Calculate \( \lim_{x\to+\infty} \frac{x^2}{e^x} \)** This is an indeterminate form \( \frac{\infty}{\infty} \), so use L'Hospital's Rule twice: First derivative: \[ \lim_{x\to\infty} \frac{2x}{e^x} \] Second derivative: \[ \lim_{x\to\infty} \frac{2}{e^x} = 0 \] **Final answer:** \(\boxed{0}\) --- ### **(c) Calculate \( \lim_{x\to-\infty} \frac{x^2}{e^x} \)** As \( x \to -\infty \), \( x^2 \to \infty \), but \( e^x \to 0^+ \), so the denominator shrinks very rapidly, making the whole fraction grow **very large and negative** (since \( e^x \) is positive but very small, and \( x^2 \) is positive). However, since for large negative \( x \), \( e^x \) dominates and approaches 0 much faster than \( x^2 \) grows, but the **sign** is positive because both numerator and denominator are positive. So, \[ \lim_{x\to-\infty} \frac{x^2}{e^x} = \infty \] --- ### **(d) Sketch of the graph** **Key features:** - Local minimum at \( x = 0 \), \( f(0) = 0 \) - Local maximum at \( x = 2 \), \( f(2) = \frac{4}{e^2} \) - As \( x \to \infty \), \( f(x) \to 0 \) - As \( x \to -\infty \), \( f(x) \to \infty \) **Sketch:** - The graph comes down from \( +\infty \) on the left, hits a minimum at \( (0,0) \), rises to a local maximum at \( (2, \frac{4}{e^2}) \), then decreases toward 0 as \( x \to +\infty \). --- ## **Summary Table** | x-value | Type | f(x) value | |---------|------------------|--------------------| | 0 | Local Minimum | 0 | | 2 | Local Maximum | \( \frac{4}{e^2} \)| | \( +\infty \) | Limit | 0 | | \( -\infty \) | Limit | \( +\infty \) | --- **Let me know if you'd like a graph sketch!**

Certainly! Based on the analysis above, here's a step-by-step guide to sketch the graph of \( f(x) = \frac{x^2}{e^x} \): ### Key features to include: 1. **Behavior at \( x \to -\infty \):** - \( f(x) \to +\infty \). The graph starts high on the far left. 2. **Critical points:** - **At \( x=0 \):** - \( f(0) = 0 \). - Local **minimum** (since \( f''(0) > 0 \)). - The graph reaches its lowest point at the origin. - **At \( x=2 \):** - \( f(2) = \frac{4}{e^2} \). - Local **maximum** (since \( f''(2) < 0 \)). - The graph peaks at this point. 3. **Behavior at \( x \to +\infty \):** - \( f(x) \to 0 \). The graph approaches the x-axis from above. ### **Sketch Instructions:** - Start from the **top left**, very high (approaching \( +\infty \)) as \( x \to -\infty \). - The curve **decreases** towards \( y=0 \) as it approaches \( x=0 \). At \( x=0 \), it reaches the **minimum point** \( (0,0) \). - After \( x=0 \), the function **rises** to a **local maximum** at \( (2, \frac{4}{e^2}) \). - Beyond \( x=2 \), the graph **decreases**, approaching \( y=0 \) again as \( x \to +\infty \). ### **Visual tips:** - Mark the critical points clearly. - Show the decreasing/increasing nature of the function between these points. - Indicate the end behavior with arrows pointing down towards the x-axis on the right and up to infinity on the left. --- ### **Summary of the graph shape:** ``` /\ / \ (peak at x=2) / \ -------/ \----------------- (approaching y=0 as x→+∞) / / (descending to +∞ as x→ -∞) ``` This should give a clear picture of how \( f(x) \) behaves over the real line. --- If you'd like, I can generate an actual plotted graph for you!

**Final Answers:** **(a)** Critical points at \( x = 0 \) and \( x = 2 \). - At \( x=0 \): \( f''(0) > 0 \) → **local minimum**. - At \( x=2 \): \( f''(2) < 0 \) → **local maximum**. **(b)** \(\displaystyle \lim_{x\to+\infty} \frac{x^2}{e^x} = 0\). **(c)** \(\displaystyle \lim_{x\to-\infty} \frac{x^2}{e^x} = +\infty\). **(d)** Graph description: - As \( x \to -\infty \), \( f(x) \to +\infty \). - Minimum at \( (0,0) \). - Maximum at \( (2, 4/e^2) \). - As \( x \to +\infty \), \( f(x) \to 0 \).