Show that $\frac{dy}{dx}=\frac{1-4 \ln x}{x^5}$.

The graph of $f$ has a horizontal tangent at point $\text{P}$. Find the coordinates of $\text{P}$.

Given that $f\text{'}\text{'}\left(x\right)=\frac{20 \ln x-9}{x^6}$, show that $\text{P}$ is a local maximum point.

Solve $f\left(x\right)>0$ for $x>0$.

Sketch the graph of $f$, showing clearly the value of the $x$-intercept and the approximate position of point $\text{P}$.

attempt to use quotient or product rule        (M1)

$\frac{dy}{dx}=\frac{x^4\left({\displaystyle \frac{1}{x}}\right)-\left(\ln x\right)\left(4x^3\right)}{{\left(x^4\right)}^2}$  OR  $\left(\ln x\right)\left(-4x^{-5}\right)+\left(x^{-4}\right)\left(\frac{1}{x}\right)$         A1

correct working         A1

$=\frac{x^3\left(1-4 \ln x\right)}{x^8}$  OR  cancelling $x^3$  OR  $\frac{-4 \ln x}{x^5}+\frac{1}{x^5}$

$=\frac{1-4 \ln x}{x^5}$         AG

[3 marks]

$f\text{'}\left(x\right)=\frac{dy}{dx}=0$        (M1)

$\frac{1-4 \ln x}{x^5}=0$

$\ln x=\frac{1}{4}$        (A1)

$x={\text{e}}^{\frac{1}{4}}$         A1

substitution of their $x$ to find $y$        (M1)

$y=\frac{\ln {\text{e}}^{{\displaystyle \frac{1}{4}}}}{{\left({\text{e}}^{\frac{1}{4}}\right)}^4}$

$=\frac{1}{4\text{e}}\left(=\frac{1}{4}{\text{e}}^{-1}\right)$         A1

$\text{P}\left({\text{e}}^{\frac{1}{4}}, \frac{1}{4\text{e}}\right)$

[5 marks]

$f\text{'}\text{'}\left({\text{e}}^{\frac{1}{4}}\right)=\frac{20 \ln {\text{e}}^{{\displaystyle \frac{1}{4}}}-9}{{\left({\text{e}}^{\frac{1}{4}}\right)}^6}$        (M1)

$=\frac{5-9}{{\text{e}}^{1.5}} \left(=-\frac{4}{{\text{e}}^{1.5}}\right)$         A1

which is negative         R1

hence $\text{P}$ is a local maximum         AG

Note: The R1 is dependent on the previous A1 being awarded.

[3 marks]

$\ln x>0$        (A1)

$x>1$        A1

[2 marks]

        A1A1A1

Note: Award A1 for one $x$-intercept only, located at $1$

     A1 for local maximum, $\text{P}$, in approximately correct position
     A1 for curve approaching $x$-axis as $x\to \infty$ (including change in concavity).

[3 marks]