91Ó°ÊÓ

To find the local extrema, we first need to find the first derivative of the given function, \(f(x) = x^{\frac{1}{x}}\). To do this, we can use the chain rule. Let's rewrite the function as $$f(x) = e^{\ln(x^{\frac{1}{x}})},$$ so the chain rule can be applied more easily. Taking the derivative with respect to \(x\), we get:$$ f'(x) = \frac{d}{dx} \, e^{\ln(x^{\frac{1}{x}})} = e^{\ln(x^{\frac{1}{x}})} \cdot \frac{d}{dx} \, \ln(x^{\frac{1}{x}}), $$since the derivative of the exponential function is itself. Next, we differentiate the argument of the exponential function, which is \(\ln(x^{\frac{1}{x}})\). To do this, we use the chain rule again and rewrite the argument as \(\frac{1}{x} \cdot \ln(x)\), so we get:$$ \frac{d}{dx} \, \ln(x^{\frac{1}{x}}) = \frac{d}{dx} \, \frac{1}{x} \cdot \ln(x) = -\frac{1}{x^2} \cdot \ln(x) + \frac{1}{x} \cdot \frac{1}{x}. $$Now, we can substitute this result back into the expression for \(f'(x)\) to obtain:$$ f'(x) = x^{\frac{1}{x}} \left(-\frac{1}{x^2} \cdot \ln(x) + \frac{1}{x^2}\right). $$

To find the critical points, we need to find the values of \(x\) such that $$f'(x) = x^{\frac{1}{x}} \left(-\frac{1}{x^2} \cdot \ln(x) + \frac{1}{x^2}\right) = 0.$$Since the term \(x^{\frac{1}{x}}\) is always positive for \(x > 0\), we only need to consider the second part of the expression:$$ -\frac{1}{x^2} \cdot \ln(x) + \frac{1}{x^2} = 0. $$Solving for \(\ln(x)\), we get that$$ \ln(x) = 1. $$Taking the exponential of both sides, we obtain$$ x = e. $$So, the critical point is \(x = e\).

To determine if the critical point corresponds to a local minimum or maximum, we need to find the second derivative of \(f(x)\) and analyze its signs. We already found the first derivative, so now we differentiate it again:$$ f''(x) = \frac{d}{dx} \, x^{\frac{1}{x}} \left(-\frac{1}{x^2} \cdot \ln(x) + \frac{1}{x^2}\right). $$Using product rule and chain rule, we obtain:$$ f''(x) = x^{\frac{1}{x}} \cdot \left(-\frac{1}{x^2} + \frac{3}{x^3} \cdot \ln(x) - \frac{2}{x^3}\right) + \left(-\frac{1}{x^2} \cdot \ln(x) + \frac{1}{x^2}\right) \cdot \frac{1}{x} \cdot x^{\frac{1}{x}-1}. $$Now, we analyze the sign of \(f''(x)\) at the critical point \(x = e\). Plugging it in, we get$$ f''(e) = e^{\frac{1}{e}} \cdot \left(-\frac{1}{e^2} + \frac{3}{e^3} \cdot \ln(e) - \frac{2}{e^3}\right) + \left(-\frac{1}{e^2} \cdot \ln(e) + \frac{1}{e^2}\right) \cdot \frac{1}{e} \cdot e^{\frac{1}{e}-1} = e^{\frac{1}{e}} \cdot \frac{1}{e^2} > 0. $$Since the second derivative is positive, this critical point corresponds to a local minimum.

To confirm our results, we need to plot the function \(f(x) = x^{\frac{1}{x}}\) and the critical point \(x = e\). Using a graphing utility, we can observe that the function indeed has a local minimum at \(x = e\), as expected. In conclusion, the function \(f(x) = x^{\frac{1}{x}}\) has a local minimum at \(x = e\).