Critical points of the function f(x)=`(x-3)^(2//3)(3x-1) `

Options

A. `(31)/(15),1`
B. `(31)/(15),3`
C. `(31)/(15),0`
D. none of these

Answer

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

### Step 1: Differentiate the function
We will use the product rule to differentiate \( f(x) \). The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative \( f'(x) \) is given by:
\[ f'(x) = u'v + uv' \]
Let:
- \( u = (x - 3)^{\frac{2}{3}} \)
- \( v = (3x - 1) \)

Now we need to find \( u' \) and \( v' \).

#### Finding \( u' \):
Using the chain rule:
\[ u' = \frac{2}{3}(x - 3)^{-\frac{1}{3}} \cdot 1 = \frac{2}{3}(x - 3)^{-\frac{1}{3}} \]

#### Finding \( v' \):
\[ v' = 3 \]

Now we can apply the product rule:
\[ f'(x) = u'v + uv' = \frac{2}{3}(x - 3)^{-\frac{1}{3}}(3x - 1) + (x - 3)^{\frac{2}{3}} \cdot 3 \]

### Step 2: Simplify \( f'(x) \)
We can rewrite \( f'(x) \):
\[ f'(x) = \frac{2(3x - 1)}{3(x - 3)^{\frac{1}{3}}} + 3(x - 3)^{\frac{2}{3}} \]

To combine these terms, we can find a common denominator:
\[ f'(x) = \frac{2(3x - 1) + 9(x - 3)}{3(x - 3)^{\frac{1}{3}}} \]

### Step 3: Set \( f'(x) = 0 \)
To find the critical points, we set \( f'(x) = 0 \):
\[ 2(3x - 1) + 9(x - 3) = 0 \]

Expanding this gives:
\[ 6x - 2 + 9x - 27 = 0 \]
\[ 15x - 29 = 0 \]
\[ x = \frac{29}{15} \]

### Step 4: Identify critical points
The critical point is \( x = \frac{29}{15} \).

### Summary of Steps
1. Differentiate the function using the product rule.
2. Simplify the derivative.
3. Set the derivative equal to zero to find critical points.
4. Solve for \( x \).