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(c) Find the critical points of \( f(x)=x^{1 / 3}(x-4) \). Sketch the graph of \( f(x) \).

Ask by Hall Hills. in Kenya
Jan 23,2025

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Answer

The critical points of \( f(x) = x^{1/3}(x - 4) \) are at \( x = 0 \) and \( x = 1 \). The graph of \( f(x) \) has a critical point at \( (0, 0) \) and another at \( (1, -3) \).

Solution

To find the **critical points** of the function \( f(x) = x^{1/3}(x - 4) \), we'll follow these steps: ### 1. Find the Derivative \( f'(x) \) First, let's differentiate \( f(x) \) with respect to \( x \): \[ f(x) = x^{1/3}(x - 4) = x^{4/3} - 4x^{1/3} \] Differentiating term by term: \[ f'(x) = \frac{4}{3}x^{1/3} - \frac{4}{3}x^{-2/3} \] ### 2. Determine Where \( f'(x) = 0 \) or \( f'(x) \) Does Not Exist **a. Setting \( f'(x) = 0 \):** \[ \frac{4}{3}x^{1/3} - \frac{4}{3}x^{-2/3} = 0 \] Multiply both sides by \( \frac{3}{4}x^{2/3} \) to eliminate the fractions and negative exponents: \[ x^{1/3} \cdot x^{2/3} - 1 = 0 \implies x - 1 = 0 \implies x = 1 \] **b. Checking Where \( f'(x) \) Does Not Exist:** The derivative \( f'(x) \) involves the term \( x^{-2/3} \), which is undefined at \( x = 0 \). Therefore, \( x = 0 \) is also a critical point. ### 3. Identify the Critical Points - **At \( x = 1 \):** \[ f(1) = 1^{1/3}(1 - 4) = 1 \times (-3) = -3 \] So, one critical point is \( (1, -3) \). - **At \( x = 0 \):** \[ f(0) = 0^{1/3}(0 - 4) = 0 \times (-4) = 0 \] Another critical point is \( (0, 0) \). ### 4. Summary of Critical Points - \( (0, 0) \) - \( (1, -3) \) ### 5. Sketch of the Graph of \( f(x) \) Here's a rough sketch of the function \( f(x) = x^{1/3}(x - 4) \): ![Graph of f(x) = x^(1/3)(x - 4)](https://i.imgur.com/your_image_link.png) **Key Features to Include in the Sketch:** 1. **Critical Points:** - Point at \( (0, 0) \): This is where the derivative does not exist. - Point at \( (1, -3) \): This is where the derivative is zero. 2. **Behavior Around Critical Points:** - **At \( x = 0 \):** The function changes its slope direction, indicating a possible inflection point. - **At \( x = 1 \):** This is likely a local minimum since the function decreases before \( x = 1 \) and increases after. 3. **End Behavior:** - As \( x \to \infty \), \( f(x) \) behaves like \( x^{4/3} \), which increases to infinity. - As \( x \to -\infty \), \( f(x) \) behaves like \( x^{4/3} \) but remains positive (since even roots yield positive values). 4. **Additional Points:** - Plot a few more points for accuracy, such as \( x = 4 \) (where \( f(4) = 4^{1/3}(0) = 0 \)). **Note:** The actual graph may require more detailed plotting for precision, especially around \( x = 0 \) due to the cube root function.

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Bonus Knowledge

To find the critical points of \( f(x) = x^{1/3}(x-4) \), we first need to compute the derivative \( f'(x) \). Using the product rule, we have: \[ f'(x) = \frac{1}{3}x^{-2/3}(x - 4) + x^{1/3}(1) = x^{-2/3}\left(\frac{1}{3}(x - 4) + x\right) = x^{-2/3}\left(\frac{4}{3}x - \frac{4}{3}\right) = \frac{4(x - 3)}{3x^{2/3}}. \] Setting \( f'(x) = 0 \) gives us \( x - 3 = 0 \), or \( x = 3 \) as a critical point. We also consider where the derivative is undefined; this occurs at \( x = 0 \). To sketch the graph of \( f(x) \), note that: - At \( x = 0 \), \( f(0) = 0 \). - At \( x = 3 \), \( f(3) = 3^{1/3}(3-4) = -\frac{3\sqrt[3]{3}}{3} = -\sqrt[3]{3} \). - At \( x = 4 \), \( f(4) = 4^{1/3}(4-4) = 0 \). The function has the following intervals for behavior analysis: - On \( (-\infty, 0) \), \( f(x) \) will approach \( 0 \). - On \( (0, 3) \), \( f(x) \) decreases and touches zero at \( (0,0) \) and reaches a negative value at \( x=3 \). - On \( (3, 4) \), \( f(x) \) increases to touch the x-axis again at \( x=4 \). Thus, the graph will showcase a curve starting from the origin, dipping below the x-axis at \( x=3 \), and rising back to touch the axis at \( x=4 \). Now you can visualize a connected and continuous graph cutting through the points identified to get an idea of how \( f(x) \) behaves overall!

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