f(x) = x2⁄3 (x − 6) 
Find the local minimum and maximum value of f. (Round your answer to two decimal places.)
local minimum value

We are given the function \( f(x) = x^{2/3}(x - 6) \), and we need to find the local minimum value.

### Steps to find local extrema:

1. **Find the first derivative \( f'(x) \):**
   To locate the critical points, we must first compute the derivative of the function and set it equal to zero.

2. **Solve \( f'(x) = 0 \) to find the critical points.**

3. **Use the second derivative test or analyze the behavior of the first derivative to determine if the critical points are local minima or maxima.**

Let’s start by finding \( f'(x) \). 

The function is given by \( f(x) = x^{2/3}(x - 6) \), and we will apply the product rule of differentiation:

\[
f'(x) = \frac{d}{dx} \left( x^{2/3} \right)(x - 6) + x^{2/3} \frac{d}{dx}(x - 6)
\]

I will compute this derivative and solve the equation for critical points.The first derivative of the function is:

\[
f'(x) = \frac{2}{3}x^{-1/3}(x - 6) + x^{2/3}
\]

Setting this equal to zero, we find that the critical point is:

\[
x = 2.40
\]

Now, we will analyze the behavior of \( f'(x) \) around this critical point to determine if it is a local minimum or maximum.

Next, I'll compute the second derivative \( f''(x) \) and use it to apply the second derivative test.The second derivative of the function is:

\[
f''(x) = -\frac{2}{9}x^{-4/3}(x - 6) + \frac{4}{3}x^{-1/3}
\]

At the critical point \( x = 2.40 \), the value of the second derivative is approximately \( 1.24 \), which is positive. Since \( f''(2.40) > 0 \), this indicates that \( x = 2.40 \) is a **local minimum**.

Now, let's calculate the local minimum value \( f(2.40) \).The local minimum value of the function \( f(x) = x^{2/3}(x - 6) \) occurs at \( x = 2.40 \), and the minimum value is approximately \( -6.45 \).

Would you like more details on any part of the process?

Here are 5 related questions to explore:
1. How do we find the local maximum of the same function?
2. What is the significance of the second derivative test in finding local extrema?
3. How can we interpret critical points that do not have a second derivative?
4. Can we find inflection points for the function?
5. How would the local extrema change if the function had a different exponent?

**Tip:** Always check the second derivative at critical points to confirm whether they are minima or maxima!

f(x) = x^(2/3)(x − 6) Find the local minimum and maximum value of f. (Round your answer to two decimal places.)

Learn how to find the local minimum of the function f(x) = x^(2/3)(x - 6) using calculus. This solution covers finding the first and second derivatives, critical points, and applying the second derivative test. Suitable for high school students in Grades 11-12.

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