f(x)=x−5x1/5


(A) Find all critical values of f
. If there are no critical values, enter -1000. If there are more than one, enter them separated by commas.
Critical value(s) = 
1


(B) Use interval notation to indicate where f(x)
 is increasing.


Note: When using interval notation in WeBWorK, you use I for ∞
, -I for −∞
, and U for the union symbol. If there are no values that satisfy the required condition, then enter "{}" without the quotation marks.

Increasing:
(-INF,1)



(C) Use interval notation to indicate where f(x)
 is decreasing.
Decreasing:
(1,INF)



(D) Find the x
-coordinates of all local maxima of f
. If there are no local maxima, enter -1000. If there are more than one, enter them separated by commas.

Local maxima at x
 = 
1


Let's break this problem into parts and verify your inputs systematically:

---

### **Given Function:**
\[ f(x) = x - 5x^{1/5} \]

#### **Step 1: Find the derivative, \( f'(x) \):**

The derivative of \( f(x) = x - 5x^{1/5} \) is:
\[
f'(x) = 1 - \frac{5}{5}x^{-4/5} = 1 - x^{-4/5}.
\]

Simplify further:
\[
f'(x) = 1 - \frac{1}{x^{4/5}}.
\]

#### **Step 2: Critical values:**

Critical values occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined.

1. **Solve \( f'(x) = 0 \):**
\[
1 - \frac{1}{x^{4/5}} = 0 \implies \frac{1}{x^{4/5}} = 1.
\]

Raise both sides to the power of \( 5/4 \):
\[
x = 1.
\]

2. **Check where \( f'(x) \) is undefined:**
\( f'(x) \) is undefined when \( x^{4/5} = 0 \), i.e., \( x = 0 \). 

However, \( x = 0 \) is not relevant here since \( x^{-4/5} \) would be undefined for negative inputs as well.

Thus, the only **critical value** is:
\[
\boxed{x = 1}.
\]

---

### **Step 3: Test intervals for increasing/decreasing behavior**

We analyze \( f'(x) = 1 - \frac{1}{x^{4/5}} \) to determine the sign of \( f'(x) \) in different intervals:

1. **Interval \( (-\infty, 1) \):**
   - Choose a test point, e.g., \( x = 0.5 \).
   - \( f'(0.5) = 1 - \frac{1}{0.5^{4/5}} \).
     - \( 0.5^{4/5} \approx 0.574 \), so \( f'(0.5) = 1 - \frac{1}{0.574} < 0 \).
   - Thus, \( f(x) \) is **decreasing** on \( (-\infty, 1) \).

2. **Interval \( (1, \infty) \):**
   - Choose a test point, e.g., \( x = 2 \).
   - \( f'(2) = 1 - \frac{1}{2^{4/5}} \).
     - \( 2^{4/5} \approx 1.319 \), so \( f'(2) = 1 - \frac{1}{1.319} > 0 \).
   - Thus, \( f(x) \) is **increasing** on \( (1, \infty) \).

---

### **Step 4: Local maxima**

From the above analysis:
- \( f(x) \) transitions from **decreasing** to **increasing** at \( x = 1 \), indicating a **local maximum** at \( x = 1 \).

---

### **Step 5: Final answers**

(A) **Critical values**:  
\[
\boxed{1}
\]

(B) **Increasing**:  
\[
\boxed{(1, \infty)}
\]

(C) **Decreasing**:  
\[
\boxed{(-\infty, 1)}
\]

(D) **Local maxima at**:  
\[
\boxed{1}
\]

---

Would you like me to verify or further explain any part of this? 😊  
Here are 5 follow-up questions for practice:  

1. How do you determine if a critical point is a maximum or minimum?
2. What does \( f'(x) > 0 \) signify about the function's behavior?
3. How do inflection points relate to second derivatives?
4. What are the steps to find global extrema on a closed interval?
5. How would the critical values change if the power of \( x \) in the given function was different?  

**Tip**: Always double-check the domain of the function to ensure no critical points are missed!

f(x) = x − 5x^(1/5). Find all critical values of f. Use interval notation to indicate where f(x) is increasing and decreasing. Find the x-coordinates of all local maxima of f.

Solve for the critical values, increasing and decreasing intervals, and local maxima of the function f(x) = x − 5x^(1/5) using calculus principles. This problem demonstrates key steps like finding derivatives, solving for critical points, and testing intervals. Suitable for students in Grades 11-12 studying calculus.

Math Problem Statement

Local maxima at x

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