Let's solve the problem step by step. **Given:** - \( f(x) = \frac{1}{x - x^2} \) - \( a = 1 \) ### **Step 1: Find \( f'(x) \)** Use the **quotient rule** or rewrite as a power: \[ f(x) = (x - x^2)^{-1} \] Now, differentiate using the chain rule: \[ f'(x) = -1 \cdot (x - x^2)^{-2} \cdot (1 - 2x) \] \[ f'(x) = \frac{-(1 - 2x)}{(x - x^2)^2} \] --- ### **Step 2: Evaluate \( f'(1) \)** Substitute \( x = 1 \): First, calculate the denominator: \[ x - x^2 = 1 - 1^2 = 0 \] So, \( f(x) \) and \( f'(x) \) are **undefined** at \( x = 1 \) because the denominator is zero. This means the function has a **vertical asymptote** at \( x = 1 \). --- ### **Step 3: Graph \( f(x) \) and the tangent line at \( x = 1 \)** - The function \( f(x) \) is undefined at \( x = 1 \). - There is **no tangent line at \( x = 1 \)** because the function is not defined there. --- ## **Final Answers** ### a. \( f'(1) \) is **undefined** because the denominator is zero at \( x = 1 \). ### b. The function \( f(x) \) and the tangent line at \( x = 1 \) **cannot be graphed** at \( x = 1 \) as the function is not defined there. --- #### **Graph Explanation** If you were to graph \( f(x) \), you'd see a vertical asymptote at \( x = 1 \) (as well as at \( x = 0 \)), and the function does not exist at those points. --- **Let me know if you'd like to see the graph of \( f(x) \) around \( x = 1 \) (excluding \( x = 1 \))!**

Let's analyze the problem carefully with detailed calculations. --- ## Given: \[ f(x) = \frac{1}{x - x^2} \] and the point \( a = 1 \). --- ## Part (a): Evaluate \( f'(a) \) ### Step 1: Rewrite \( f(x) \) Express \( f(x) \) as: \[ f(x) = (x - x^2)^{-1} \] ### Step 2: Find \( f'(x) \) using the chain rule Let \( u = x - x^2 \). Then, \[ f(x) = u^{-1} \] Derivative: \[ f'(x) = -1 \cdot u^{-2} \cdot u' \] Calculate \( u' \): \[ u' = \frac{d}{dx}(x - x^2) = 1 - 2x \] Substitute back: \[ f'(x) = - (1 - 2x) \cdot (x - x^2)^{-2} \] or, \[ f'(x) = \frac{-(1 - 2x)}{(x - x^2)^2} \] ### **Summary for part (a):** \[ \boxed{ f'(x) = \frac{-(1 - 2x)}{(x - x^2)^2} } \] --- ## Part (b): Evaluate the derivative at \( x = a = 1 \) Substitute \( x = 1 \): \[ f'(1) = \frac{-(1 - 2 \times 1)}{(1 - 1^2)^2} = \frac{-(1 - 2)}{(1 - 1)^2} = \frac{-(-1)}{0^2} = \frac{1}{0} \] Since the denominator is zero, \( f'(1) \) is **undefined** (tends to infinity). ### **Interpretation:** - The derivative does **not exist** at \( x=1 \). - The function \( f(x) \) is **not defined** at \( x=1 \) because the denominator \( x - x^2 \) is zero there. - The point \( x=1 \) corresponds to a **vertical asymptote**. --- ## **Final answers:** ### a. \( f'(1) \) is **undefined** (the derivative tends to infinity or does not exist). ### b. **Graph of \( f(x) \) and tangent line at \( x=1 \):** - Since \( f(x) \) is undefined at \( x=1 \), **there is no tangent line** at \( x=1 \). - The function has a **vertical asymptote** at \( x=1 \), meaning the graph approaches infinity or negative infinity as \( x \to 1 \). --- ## **Summary:** | Part | Result | Explanation | |---------|---------|--------------| | (a) | \( f'(1) \) is undefined | The denominator in the derivative is zero at \( x=1 \). | | (b) | No tangent line at \( x=1 \) | The function is not defined there; hence, no tangent line exists. | --- **Note:** If you want to analyze the behavior of the function near \( x=1 \), you can examine limits as \( x \to 1^{\pm} \): \[ \lim_{x \to 1^-} f(x) = -\infty, \quad \lim_{x \to 1^+} f(x) = +\infty \] indicating a vertical asymptote at \( x=1 \).

Certainly! While I can't produce images directly here, I can guide you step-by-step on how to sketch the graph of \(f(x) = \frac{1}{x - x^2}\) around \(x=1\), highlighting key features. --- ### Step-by-step guide to sketch the graph of \(f(x) = \frac{1}{x - x^2}\) --- ### 1. **Rewrite the function** \[ f(x) = \frac{1}{x - x^2} = \frac{1}{x(1 - x)} \] This form shows potential asymptotes where the denominator is zero: - \(x = 0\) - \(x = 1\) --- ### 2. **Identify asymptotes** - **Vertical asymptotes** at \(x=0\) and \(x=1\): - As \(x \to 0^-\), \(f(x) \to -\infty\) - As \(x \to 0^+\), \(f(x) \to +\infty\) - As \(x \to 1^-\), \(f(x) \to -\infty\) - As \(x \to 1^+\), \(f(x) \to +\infty\) - **Horizontal asymptote**: Since degree of numerator < degree of denominator (constant over quadratic), \[ \lim_{x \to \pm \infty} f(x) = 0 \] So, \(y=0\) is a horizontal asymptote. --- ### 3. **Analyze the behavior in different intervals** - For \(x < 0\): - Pick \(x = -1\): \[ f(-1) = \frac{1}{-1 - 1} = \frac{1}{-2} = -0.5 \] - As \(x \to 0^-\), \(f(x) \to -\infty\). - For \(0 < x < 1\): - Pick \(x=0.5\): \[ f(0.5) = \frac{1}{0.5 - 0.25} = \frac{1}{0.25} = 4 \] - As \(x \to 0^+\), \(f(x) \to +\infty\). - As \(x \to 1^-\), \(f(x) \to -\infty\). - For \(x > 1\): - Pick \(x=2\): \[ f(2) = \frac{1}{2 - 4} = \frac{1}{-2} = -0.5 \] - As \(x \to 1^+\), \(f(x) \to +\infty\). - As \(x \to \infty\), \(f(x) \to 0^-\). --- ### 4. **Plot key points and asymptotes** - Draw vertical dashed lines at \(x=0\) and \(x=1\) to indicate asymptotes. - Mark points: - \((-1, -0.5)\) - \((0.5, 4)\) - \((2, -0.5)\) - The graph approaches infinity near \(x=0^+\), \(x=1^-\), and \(x=1^+\). --- ### 5. **Sketch the graph** - From \(-\infty\) to \(0\): - The curve goes from \(-\infty\) (near \(x=0^-\)) up to about \(-0.5\) at \(x=-1\). - It approaches the x-axis from below as \(x \to -\infty\). - From just after \(0\) to just before \(1\): - The curve jumps from \(+\infty\) at \(x \to 0^+\) down to \(-\infty\) as \(x \to 1^-\). - At \(x=0.5\), it reaches a value of 4. - From just after \(1\) to \(\infty\): - The graph comes from \(+\infty\) at \(x=1^+\), then decreases toward 0 as \(x \to \infty\). - At \(x=2\), the value is \(-0.5\). --- ### **Summary of the sketch:** - **Vertical asymptotes at \(x=0\) and \(x=1\).** - **Horizontal asymptote at \(y=0\).** - The function is negative on \((-\infty, 0)\), positive on \((0,1)\), negative again on \((1,\infty)\). - The graph "blows up" near the asymptotes and approaches zero far away on both ends. --- ### **Note:** If you can use graphing tools (like Desmos, GeoGebra), plot: \[ f(x) = \frac{1}{x(1 - x)} \] and observe the asymptotic behaviors described. --- **Would you like me to help interpret specific features further or provide a sketch outline?**

Certainly! While I can't directly draw images here, I can provide a detailed ASCII-style sketch to help visualize the graph of \(f(x) = \frac{1}{x - x^2} = \frac{1}{x(1 - x)}\), along with the key features such as asymptotes and behavior in different regions. --- ### **ASCII Sketch of \(f(x) = \frac{1}{x(1 - x)}\)** ``` y | /\ | / \ | / \ | / \ | / \ |---------------------------/----------\------------------- x | / \ | / \ | / \ | / \ | / \ | ---|---------|------------------------|---------|--- | 0 1 ... ``` --- ### **Features explained:** - **Vertical asymptotes** at: - \(x=0\) (left vertical dashed line) - \(x=1\) (right vertical dashed line) - **Behavior near asymptotes:** - As \(x \to 0^-\), \(f(x) \to -\infty\) - As \(x \to 0^+\), \(f(x) \to +\infty\) - As \(x \to 1^-\), \(f(x) \to -\infty\) - As \(x \to 1^+\), \(f(x) \to +\infty\) - **Horizontal asymptote** at \(y=0\), as the graph approaches zero for large \(|x|\). - **Behavior in regions:** - \(x < 0\): The graph is negative, approaching \(-\infty\) near 0, then dips down and approaches 0 from below as \(x \to -\infty\). - \(0 < x < 1\): The graph is positive, starting from \(+\infty\) near 0, reaching a maximum (e.g., at \(x=0.5\), \(f=4\)), then dropping to \(-\infty\) as \(x \to 1^-\). - \(x > 1\): The graph is negative, starting from \(+\infty\) at \(x=1^+\), then decreasing toward 0 as \(x \to \infty\). --- ### **Summary:** - The graph has **two vertical asymptotes at \(x=0\) and \(x=1\)**. - It crosses the y-axis at \(x=0\) with a vertical asymptote. - Approaches zero as \(x \to \pm \infty\). - The function is positive between 0 and 1, negative elsewhere. - The general shape resembles two branches approaching asymptotes and the x-axis. --- ### **How to Sketch It Yourself:** 1. Draw vertical dashed lines at \(x=0\) and \(x=1\). 2. Mark key points: - Near \(x=0^+\), \(f \to +\infty\), - At \(x=0.5\), \(f=4\), - Near \(x=1^-\), \(f \to -\infty\), - Near \(x=1^+\), \(f \to +\infty\), - At \(x=2\), \(f=-0.5\), - As \(x \to \pm \infty\), \(f \to 0\). 3. Sketch the branches respecting asymptotes and these points. --- If you prefer, you can use graphing tools like Desmos or GeoGebra to input: \[ f(x) = \frac{1}{x(1 - x)} \] and see a precise graph with all features. --- Let me know if you'd like further clarification or a step-by-step guide to plotting!