To find the critical points of the function \( f(x) = 2x + 3(x)^{\frac{2}{3}} \), we first calculate its derivative. The function is defined for all real numbers. We will examine its first and second derivatives to identify local maxima and minima.
\(f'(x) = \frac{d}{dx} [2x + 3(x)^{\frac{2}{3}}] = 2 + 3 \cdot \frac{2}{3}(x)^{-\frac{1}{3}} = 2 + 2(x)^{-\frac{1}{3}}\)
\(2 + 2(x)^{-\frac{1}{3}} = 0\)
\((x)^{-\frac{1}{3}} = -1 \Rightarrow x^{-\frac{1}{3}} = -1 \Rightarrow x = (-1)^{-3} = -1\)
\(f''(x) = \frac{d}{dx} [2 + 2(x)^{-\frac{1}{3}}] = 0 - \frac{2}{3}(x)^{-\frac{4}{3}}\)
\(f''(-1) = - \frac{2}{3}(-1)^{-\frac{4}{3}} = -\frac{2}{3}(-1)^{4/3} = -\frac{2}{3}(1) = -\frac{2}{3}\)
Since \( f''(-1) \lt 0 \), the function has a local maximum at \( x = -1 \).
\(f'(x) = 2 + \frac{2}{x^{1/3}}\). As \( x \to 0^+ \), \( f'(x) \) becomes very large.
This indicates that the function increases as it approaches \( x = 0 \). Since \(f'\) changes sign around \(x=0\), there is a minimum near \(x=0\).
The function \( f(x) \) exhibits exactly one local maximum and exactly one local minimum.