Consider the following.
f(x) = 5x − 15x1/3
Find 
f '(x).
f '(x) =    
Correct: Your answer is correct.
Find all critical numbers of 
f(x).
 (Enter your answers as a comma-separated list.)
x =    
Incorrect: Your answer is incorrect.
Find all relative extrema of 
f(x).
 (If an answer does not exist, enter DNE.)
relative maximum    	(x, y)	 = 	

  
Incorrect: Your answer is incorrect.
 

 
relative minimum    	(x, y)	 = 	

  
Correct: Your answer is correct.
 


Let's go through each part of the problem step by step to make sure we can solve it properly.

### Step 1: Finding the Derivative \( f'(x) \)

The function is given as:
\[
f(x) = 5x - 15x^{1/3}
\]
We will now differentiate \( f(x) \) using basic differentiation rules:

- The derivative of \( 5x \) is \( 5 \).
- The derivative of \( -15x^{1/3} \) uses the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \). Here \( n = \frac{1}{3} \), so the derivative is:
  \[
  \frac{d}{dx} \left( -15x^{1/3} \right) = -15 \times \frac{1}{3} x^{\frac{1}{3} - 1} = -5x^{-2/3}
  \]

Thus, the derivative \( f'(x) \) is:
\[
f'(x) = 5 - 5x^{-2/3}
\]
This is the first derivative.

### Step 2: Finding the Critical Numbers

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

Set \( f'(x) = 0 \) to find where the slope is zero:
\[
5 - 5x^{-2/3} = 0
\]
Solve for \( x \):
\[
5 = 5x^{-2/3}
\]
\[
1 = x^{-2/3}
\]
\[
x^{2/3} = 1
\]
Now raise both sides to the power of \( 3/2 \) to solve for \( x \):
\[
x = 1
\]

To find where \( f'(x) \) is undefined, notice that \( f'(x) = 5 - 5x^{-2/3} \) is undefined when \( x^{-2/3} \) is undefined, which happens when \( x = 0 \).

Thus, the critical numbers are:
\[
x = 0, 1
\]

### Step 3: Finding the Relative Extrema

To determine whether the critical numbers represent relative maxima or minima, we can use the second derivative test or analyze the sign of \( f'(x) \) around the critical numbers.

Let's test the sign of \( f'(x) \) around \( x = 1 \) and \( x = 0 \).

- For \( x = 0 \), plug values slightly less than and greater than 0 into \( f'(x) \) to determine the behavior of the function around 0.
- For \( x = 1 \), we can do the same.

Finally, we can calculate the value of \( f(x) \) at those points to get the relative extrema.

---

Would you like me to proceed with evaluating the extrema, or do you have any other questions so far? Here are five follow-up questions you might consider:

1. How do you find critical points from a function?
2. What is the difference between a relative maximum and minimum?
3. How does the second derivative test work?
4. What does the power rule state for derivatives?
5. Can a critical point also be a point where the derivative is undefined?

**Tip:** When solving for critical points, always check where the derivative is zero **and** where it is undefined.

Consider the following. f(x) = 5x − 15x^(1/3) Find f '(x). f '(x) = Find all critical numbers of f(x). (Enter your answers as a comma-separated list.) Find all relative extrema of f(x).

Learn how to find the derivative of f(x) = 5x - 15x^(1/3), determine critical numbers, and identify relative extrema. This calculus problem involves the power rule, critical points, and extrema analysis, suitable for students in Grades 11-12 and college-level calculus.

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