Find the critical points of `f(x)=x^(2//3)(2x-1)`

Answer

To find the critical points of the function \( f(x) = x^{2/3}(2x - 1) \), we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \)

We will use the product rule to differentiate \( f(x) \). The product rule states that if \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

Let:
- \( u(x) = x^{2/3} \)
- \( v(x) = 2x - 1 \)

Now, we need to find \( u'(x) \) and \( v'(x) \):

1. **Differentiate \( u(x) \)**:
\[ u'(x) = \frac{2}{3}x^{-1/3} \]

2. **Differentiate \( v(x) \)**:
\[ v'(x) = 2 \]

Now apply the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
Substituting the derivatives:
\[ f'(x) = \left(\frac{2}{3}x^{-1/3}\right)(2x - 1) + (x^{2/3})(2) \]

### Step 2: Simplify \( f'(x) \)

Now, we simplify the expression:
\[ f'(x) = \frac{2}{3}x^{-1/3}(2x - 1) + 2x^{2/3} \]

To combine these terms, we will find a common denominator:
\[ f'(x) = \frac{2(2x - 1)}{3x^{1/3}} + \frac{6x^{2/3}}{3x^{1/3}} \]
\[ = \frac{2(2x - 1) + 6x}{3x^{1/3}} \]
\[ = \frac{4x - 2 + 6x}{3x^{1/3}} = \frac{10x - 2}{3x^{1/3}} \]

### Step 3: Set the derivative equal to zero

To find the critical points, we set \( f'(x) = 0 \):
\[ \frac{10x - 2}{3x^{1/3}} = 0 \]
This implies:
\[ 10x - 2 = 0 \]
Solving for \( x \):
\[ 10x = 2 \quad \Rightarrow \quad x = \frac{2}{10} = \frac{1}{5} \]

### Step 4: Identify where the derivative does not exist

Next, we check where \( f'(x) \) does not exist. The derivative \( f'(x) \) is undefined when the denominator is zero:
\[ 3x^{1/3} = 0 \quad \Rightarrow \quad x^{1/3} = 0 \quad \Rightarrow \quad x = 0 \]

### Conclusion

The critical points of the function \( f(x) = x^{2/3}(2x - 1) \) are:
1. \( x = \frac{1}{5} \) (where \( f'(x) = 0 \))
2. \( x = 0 \) (where \( f'(x) \) does not exist)