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  • Fill in the blanks in each of the following The least value of the function f(x) = ax +b/x (a > 0, b > 0, x > 0) is ______.

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 ______.

Answers (1)

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 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 quotient rule on 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 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 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 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 function f(x)=a x+\frac{b}{x}(a>0, b>0, x>0)$ is 2 \sqrt{a b}$

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