+26367 Find the minimum value of \(f(x) = x + \dfrac{4}{x}\) for \(x > 0\).
\(\begin{array}{|rcll|} \hline f(x) &=& x+ \dfrac{4}{x} \\\\ f(x) &=& x+ 4x^{-1} \quad | \quad \text{minimum value, if } f'(x) = 0 \\ f'(x) &=& 1+ 4(-1)x^{-2} \\ f'(x) &=& 1- \dfrac{4}{x^2} \\ 0 &=& 1- \dfrac{4}{x^2} \\ \dfrac{4}{x^2} &=& 1 \\ x^2 &=& 4 \\ x &=& \pm 2 \\ x &=& +2 \quad | \quad x > 0 \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline f(x) &=& x+ \dfrac{4}{x} \quad | \quad x=2 \\\\ f(2) &=& 2+ \dfrac{4}{2} \\\\ f(2) &=& 2+ 2 \\\\ \mathbf{f(2)} &=& \mathbf{4} \\ \hline \end{array}\)
The minimum value of f(x) is 4
