let (f) be the function defined by (f(x)=ksqrt{x}-ln x) for (x > 0), where (k) is a positive constant. (a)…

Explanation:

Step1: Find the first - derivative

Use the power - rule and the derivative of natural logarithm. The derivative of $\sqrt{x}=x^{\frac{1}{2}}$ is $\frac{1}{2}x^{-\frac{1}{2}}$ and the derivative of $\ln x$ is $\frac{1}{x}$. So, $f(x)=k\sqrt{x}-\ln x = kx^{\frac{1}{2}}-\ln x$, and $f^{\prime}(x)=\frac{k}{2\sqrt{x}}-\frac{1}{x}=\frac{k\sqrt{x}-2}{2x}$.

Step2: Find the second - derivative

Differentiate $f^{\prime}(x)$ with respect to $x$. Using the quotient - rule $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}$, where $u = k\sqrt{x}-2=\frac{k}{\sqrt{x}}-2$ and $v = 2x$. $u^{\prime}=\frac{k}{2\sqrt{x}}$ and $v^{\prime}=2$. Then $f^{\prime\prime}(x)=\frac{\frac{k}{2\sqrt{x}}\cdot2x-(k\sqrt{x}-2)\cdot2}{(2x)^{2}}=\frac{k\sqrt{x}-2k\sqrt{x} + 4}{4x^{2}}=\frac{4 - k\sqrt{x}}{4x^{2}}$.

Step3: Find the value of $k$ for the critical point

A critical point occurs where $f^{\prime}(x)=0$. Set $x = 1$ in $f^{\prime}(x)$: $f^{\prime}(1)=\frac{k\sqrt{1}-2}{2\times1}=0$. Then $k - 2=0$, so $k = 2$.

Step4: Determine the nature of the critical point

Substitute $k = 2$ into $f^{\prime\prime}(x)$. $f^{\prime\prime}(x)=\frac{4 - 2\sqrt{x}}{4x^{2}}$. When $x = 1$, $f^{\prime\prime}(1)=\frac{4-2}{4}=\frac{1}{2}>0$. Since $f^{\prime}(1) = 0$ and $f^{\prime\prime}(1)>0$, $f(x)$ has a relative minimum at $x = 1$.

Step5: Find the value of $k$ for the inflection point on the $x$ - axis

An inflection point occurs where $f^{\prime\prime}(x)=0$, so $\frac{4 - k\sqrt{x}}{4x^{2}}=0$, which gives $4 - k\sqrt{x}=0$, or $k\sqrt{x}=4$. At an inflection point on the $x$ - axis, $f(x)=0$. Since $f(x)=k\sqrt{x}-\ln x$ and $k\sqrt{x}=4$, then $4-\ln x=0$, so $\ln x = 4$ and $x = e^{4}$. Substitute $x = e^{4}$ into $k\sqrt{x}=4$, we get $k\sqrt{e^{4}}=4$, $ke^{2}=4$, so $k=\frac{4}{e^{2}}$.

Answer:

(a) $f^{\prime}(x)=\frac{k\sqrt{x}-2}{2x}$, $f^{\prime\prime}(x)=\frac{4 - k\sqrt{x}}{4x^{2}}$ (b) $k = 2$, $f(x)$ has a relative minimum at $x = 1$ (c) $k=\frac{4}{e^{2}}$