Question

The function \( f(x) = 2x + 3(x)^{\frac{2}{3}}, x \in \mathbb{R} \), has

• A. exactly one point of local minima and no point of local maxima
• B. exactly one point of local maxima and no point of local minima
• C. exactly one point of local maxima and exactly one point of local minima
• D. exactly two points of local maxima and exactly one point of local minima

Answer 1

The Correct Option is C

To determine the critical points of the function \( f(x) = 2x + 3(x)^{\frac{2}{3}} \), we start by finding its derivative. The function is defined for all real numbers, and we need to analyze its first and second derivatives to identify points of local maxima and minima.

1. Find the first derivative of \( f(x) \): 

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

1. Set the first derivative equal to zero to find the critical points:

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

1. Simplify and solve for \( x \):

\((x)^{-\frac{1}{3}} = -1 \Rightarrow x^{-\frac{1}{3}} = -1 \Rightarrow x = (-1)^{-3} = -1\)

1. Find the second derivative to determine the nature of critical points:

\(f''(x) = \frac{d}{dx} [2 + 2(x)^{-\frac{1}{3}}] = 0 - \frac{2}{3}(x)^{-\frac{4}{3}}\)

1. Evaluate the second derivative at \( x = -1 \):

\(f''(-1) = - \frac{2}{3}(-1)^{-\frac{4}{3}} = -\frac{2}{3}(-1)^{4/3} = -\frac{2}{3}(1) = -\frac{2}{3}\)

1. Determine the nature of the critical point:

Since \( f''(-1) \lt 0 \), the function has a local maximum at \( x = -1 \).

1. Investigate further points:
2. Evaluate the behavior of the function as \( x \to 0 \) from the positive side (as the derivative involves negative powers).

\(f'(x) = 2 + \frac{2}{x^{1/3}}\) which becomes very large as \( x \to 0^+ \).

This indicates that the function is increasing towards \( x = 0 \). As \(-\infty\to 0\), \(f'\) changes sign around \(x=0\), suggesting a minimum near there.

1. Thus, we have:
   • A local maximum at \(x = -1\).
   • A local minimum at \(x \rightarrow 0^+\).

Based on this analysis, the given function \( f(x) \) has exactly one point of local maxima and exactly one point of local minima.

Answer 2

Solution: To find the local maxima and minima, we calculate the derivative \( f'(x) \) and analyze the critical points.

Step 1. Finding \( f'(x)\):
    \(f(x) = 2x + 3(x)^{\frac{1}{3}}\)
  \(f'(x) = 2 + 2x^{-\frac{2}{3}}\)
  \(= 2 \left( 1 + \frac{1}{x^{\frac{2}{3}}} \right)\)

Step 2. Setting \( f'(x) = 0 \) to find critical points:
\(2 \left( 1 + \frac{1}{x^{\frac{2}{3}}} \right) = 0\)
  \(1 + \frac{1}{x^{\frac{2}{3}}} = 0\)  
  \(x^{\frac{2}{3}} = -1 \implies x = -1\)

Step 3. Analyzing the sign of \( f'(x) \) around the critical points \( x = -1 \) and \( x = 0 \):

So, the function has a local maximum (M) at \( x = -1 \) and a local minimum (m) at \( x = 0 \).  

The Correct Answer is: Exactly one point of local maxima and exactly one point of local minima