The number of critical points of the function `f(x) = (2x – 3)^(2//3) (2x + 1)` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

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

### Step 1: Differentiate the function
We will use the product rule for differentiation. The product rule states that if \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

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

Now we differentiate \( u(x) \) and \( v(x) \):

1. **Differentiate \( u(x) \)**:
\[ u'(x) = \frac{d}{dx}[(2x - 3)^{2/3}] = \frac{2}{3}(2x - 3)^{-1/3} \cdot 2 = \frac{4}{3}(2x - 3)^{-1/3} \]

2. **Differentiate \( v(x) \)**:
\[ v'(x) = \frac{d}{dx}(2x + 1) = 2 \]

### Step 2: Apply the product rule
Now we apply the product rule:
\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]
Substituting the derivatives we found:
\[ f'(x) = \left(\frac{4}{3}(2x - 3)^{-1/3}\right)(2x + 1) + (2x - 3)^{2/3}(2) \]

### Step 3: Simplify the derivative
Now we simplify \( f'(x) \):
\[ f'(x) = \frac{4(2x + 1)}{3(2x - 3)^{1/3}} + 2(2x - 3)^{2/3} \]
To combine these terms, we can find a common denominator:
\[ f'(x) = \frac{4(2x + 1) + 6(2x - 3)^{2/3}(2x - 3)^{1/3}}{3(2x - 3)^{1/3}} \]

### Step 4: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 4(2x + 1) + 6(2x - 3) = 0 \]
Solving this gives:
\[ 4(2x + 1) = -6(2x - 3) \]
Expanding both sides:
\[ 8x + 4 = -12x + 18 \]
Combining like terms:
\[ 20x = 14 \quad \Rightarrow \quad x = \frac{14}{20} = \frac{7}{10} \]

### Step 5: Check where the derivative is undefined
Next, we check where the derivative is undefined, which occurs when the denominator is zero:
\[ (2x - 3)^{1/3} = 0 \quad \Rightarrow \quad 2x - 3 = 0 \quad \Rightarrow \quad x = \frac{3}{2} \]

### Conclusion
Thus, we have two critical points:
1. \( x = \frac{7}{10} \)
2. \( x = \frac{3}{2} \)

Therefore, the number of critical points of the function \( f(x) \) is **2**.

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