Fill in the blanks in each of the following The least value of the function $f(x)=ax+\frac{b}{x}$ $(a > 0, b > 0, x > 0)$ is ______.

Given  $f(x)=ax+\frac{b}{x}$

After applying the derivative

$f^{\prime}(x)=\frac{d\left(a x+\frac{b}{x}\right)}{d x}$
Apply the sum rule and get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(\mathrm{ax})}{\mathrm{dx}}+\frac{\mathrm{d}\left(\frac{\mathrm{b}}{\mathrm{x}}\right)}{\mathrm{dx}}$
Apply the quotient rule on the second part and get
$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{a}+\frac{\mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{b})}{\mathrm{d} \mathrm{x}}-\mathrm{b} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{d} \mathrm{x}}}{(\mathrm{x})^{2}} \\f^{\prime}(x)=a+\frac{x \cdot 0-b \cdot 1}{x^{2}} \\f^{\prime}(x)=a-\frac{b}{x^{2}}$

Equate it with 0 and get
$\\f^{\prime}(x)=0 \\\Rightarrow \mathrm{a}-\frac{\mathrm{b}}{\mathrm{x}^{2}}=0 \\\Rightarrow \mathrm{a}=\frac{\mathrm{b}}{\mathrm{x}^{2}} \\\Rightarrow x^{2}=\frac{b}{a} \\\Rightarrow x=\sqrt{\frac{b}{a}}( as x>0)$
Now given by the second derivative,
Apply derivative and get
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}\left(\mathrm{a}-\frac{\mathrm{b}}{\mathrm{x}^{2}}\right)}{\mathrm{dx}}$
Apply the sum rule and get
$\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(\mathrm{a})}{\mathrm{dx}}-\frac{\mathrm{d}\left(\frac{\mathrm{b}}{\mathrm{x}^{2}}\right)}{\mathrm{dx}}$

Apply the quotient rule on the second part and get
$\\ \mathrm{f}^{\prime \prime}(\mathrm{x})=0-\frac{\mathrm{x}^{2} \cdot \frac{\mathrm{d}(\mathrm{b})}{\mathrm{d} \mathrm{x}}-\mathrm{b} \cdot \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}}{\left(\mathrm{x}^{2}\right)^{2}} \\\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{x}^{2} \cdot 0-\mathrm{b} \cdot(2 \mathrm{x})}{\mathrm{x}^{4}} \\f^{\prime \prime}(x)=\frac{2 x b}{x^{4}}=\frac{2 b}{x^{3}}$
Now, equate it with $x=\sqrt{\frac{b}{a}}$
$\\\left(\mathrm{f}^{\prime \prime}(\mathrm{x})\right) \mathrm{x}=\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}=\frac{2 \mathrm{~b}}{\mathrm{x}^{3}} \\\left(\mathrm{f}^{\prime \prime}(\mathrm{x})\right)_{\mathrm{x}=\sqrt{\frac{\mathrm{b}}{2}}}=\frac{2 \mathrm{~b}}{\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)^{3}} \\\left(\mathrm{f}^{\prime \prime}(\mathrm{x})\right)_{\mathrm{x}=\sqrt{\frac{\mathrm{b}}{2}}}=\frac{2 \mathrm{ab}}{\mathrm{b}}=2 \mathrm{a}>0$

The least value of f(x) is

$\begin{array}{l} \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\mathrm{ax}+\frac{\mathrm{b}}{\mathrm{x}} \\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\mathrm{a}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)+\frac{\mathrm{b}}{\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)} \\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\mathrm{a}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)+\mathrm{b} \sqrt{\frac{\mathrm{a}}{\mathrm{b}}} \\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{\mathrm{ab}+\mathrm{ab}}{\sqrt{\mathrm{ab}}} \\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{2 \mathrm{ab}}{\sqrt{\mathrm{ab}}} \end{array}$

Multiply and divide by $\sqrt{\mathrm{ab}}$ and get
$\\ \Rightarrow \mathrm{f}\left(\sqrt{\frac{\mathrm{b}}{\mathrm{a}}}\right)=\frac{2 \mathrm{ab}}{\sqrt{\mathrm{ab}}} \times \frac{\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{ab}}} \\\Rightarrow f\left(\sqrt{\frac{b}{a}}\right)=2 \sqrt{a b}$
Therefore, the least value of the function $f(x)=a x+\frac{b}{x}(a>0, b>0, x>0)$ is $2 \sqrt{a b}$