Consider function `f(x)=2x+3x^(2/3)`

Options

A. Exactly 1 local minima & no local maxima
B. Exactly 1 local minima & 1 local maxima
C. No local minima & 1 local maxima
D. No local maxima & 1 local minima

Answer

To solve the problem involving the function \( f(x) = 2x + 3x^{2/3} \) and to find the cases for maxima and minima, we will follow these steps:

### Step 1: Find the first derivative
To find the critical points, we first need to compute the first derivative of the function \( f(x) \).

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

### Step 2: Set the first derivative to zero
Next, we set the first derivative equal to zero to find the critical points.

\[ 2 + 2x^{-1/3} = 0 \]

Rearranging gives:

\[ 2x^{-1/3} = -2 \]

Dividing both sides by 2:

\[ x^{-1/3} = -1 \]

Taking the reciprocal gives:

\[ x^{1/3} = -1 \implies x = -1 \]

### Step 3: Check for other critical points
We also need to consider the point where the derivative is undefined, which occurs when \( x = 0 \) (since \( x^{-1/3} \) is undefined at \( x = 0 \)).

Thus, the critical points are \( x = -1 \) and \( x = 0 \).

### Step 4: Analyze the sign of the first derivative
To determine whether these critical points are maxima or minima, we analyze the sign of \( f'(x) \) around these points.

1. For \( x < -1 \): Choose \( x = -2 \)
\[ f'(-2) = 2 + 2(-2)^{-1/3} = 2 + 2(-\frac{1}{\sqrt[3]{2}}) > 0 \quad (\text{since } -\frac{1}{\sqrt[3]{2}} \text{ is negative}) \]

2. For \( -1 < x < 0 \): Choose \( x = -0.5 \)
\[ f'(-0.5) = 2 + 2(-0.5)^{-1/3} < 0 \quad (\text{since } (-0.5)^{-1/3} \text{ is negative}) \]

3. For \( x > 0 \): Choose \( x = 1 \)
\[ f'(1) = 2 + 2(1)^{-1/3} = 4 > 0 \]

### Step 5: Determine maxima and minima
From the sign analysis:
- At \( x = -1 \), \( f' \) changes from positive to negative, indicating a local maximum.
- At \( x = 0 \), \( f' \) changes from negative to positive, indicating a local minimum.

### Conclusion
Thus, we have found:
- One local maximum at \( x = -1 \)
- One local minimum at \( x = 0 \)

### Final Answer
The function \( f(x) = 2x + 3x^{2/3} \) has exactly one local maximum and one local minimum.

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