Let exp(x) denote the exponential function e^ | vedclass

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(C) Given $f(x) = \exp\left(x^{\frac{1}{x}}\right)$. Let $g(x) = x^{\frac{1}{x}}$. Since $\exp(u)$ is an increasing function, the minimum of $f(x)$ oc

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$\exp\left(5^{\frac{1}{5}}\right)$

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(C) Given $f(x) = \exp\left(x^{\frac{1}{x}}\right)$. Let $g(x) = x^{\frac{1}{x}}$. Since $\exp(u)$ is an increasing function, the minimum of $f(x)$ occurs where $g(x)$ is minimum.Taking the natural logarithm of $g(x)$, we have $\ln(g(x)) = \frac{\ln x}{x}$.Differentiating with respect to $x$: $\frac{g'(x)}{g(x)} = \frac{x(\frac{1}{x}) - \ln x(1)}{x^2} = \frac{1 - \ln x}{x^2}$.Setting $g'(x) = 0$ gives $1 - \ln x = 0$, so $x = e \approx 2.718$.For $x  0$ (increasing), and for $x > e$, $g'(x) Since $e \in [2, 5]$, the function $g(x)$ increases on $[2, e]$ and decreases on $[e, 5]$.The minimum value of $g(x)$ on $[2, 5]$ must occur at the endpoints $x = 2$ or $x = 5$.Comparing $g(2) = 2^{1/2} = \sqrt{2} \approx 1.414$ and $g(5) = 5^{1/5} \approx 1.3797$.Since $g(5) Thus, the minimum value of $f(x)$ is $\exp\left(5^{\frac{1}{5}}\right)$.

Let $\exp(x)$ denote the exponential function $e^x$. If $f(x) = \exp\left(x^{\frac{1}{x}}\right)$ for $x > 0$, then the minimum value of $f$ in the interval $[2, 5]$ is:

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