The critical points of the function `f(x) = (x +2)^(2//3) (2x-1)` are

Options

A. `-1 and 2`
B. 1
C. `1 and -1//2`
D. `-1 and -2`

Answer

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

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

We will use the product rule for differentiation, which states that if \( f(x) = u(x)v(x) \), then:

\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]

Here, let \( u(x) = (x + 2)^{\frac{2}{3}} \) and \( v(x) = (2x - 1) \).

### Step 2: Differentiate \( u(x) \) and \( v(x) \)

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

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

### Step 3: Apply the product rule

Now, substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule:

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

### Step 4: Simplify \( f'(x) \)

Combine the terms:

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

To combine these fractions, we can express \( 2(x + 2)^{\frac{2}{3}} \) with a common denominator:

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

### Step 5: Set \( f'(x) = 0 \) and find critical points

Setting the numerator equal to zero:

\[ 2(2x - 1) + 6(x + 2) = 0 \]

Expanding and simplifying:

\[ 4x - 2 + 6x + 12 = 0 \implies 10x + 10 = 0 \implies x = -1 \]

### Step 6: Determine where \( f'(x) \) is not defined

The derivative \( f'(x) \) is not defined when the denominator is zero:

\[ 3(x + 2)^{\frac{1}{3}} = 0 \implies x + 2 = 0 \implies x = -2 \]

### Conclusion

The critical points of the function \( f(x) \) are:

\[ \text{Critical Points: } x = -1 \text{ and } x = -2 \]