Derivative, maximum point, positive values and sketch of a logarithmic function divided by x squared
Derivative, maximum point, positive values and sketch of a logarithmic function divided by x squared
16 marks
The function \(f\) is defined by \(f(x)=\dfrac{1+\ln x}{x^2}\) for \(x>0\).
- Show that \(f'(x)=-\dfrac{1+2\ln x}{x^3}\).[3]
- The tangent to the graph of \(f\) at the point \(P\) is parallel to the \(x\)-axis. Find the coordinates of \(P\), giving your answers in exact form.[5]
- It is given that \(f''(x)=\dfrac{1+6\ln x}{x^4}\). Hence show that the stationary point at \(P\) is a maximum.[3]
- Find the set of values of \(x\) for which \(f(x)>0\).[2]
- Sketch the graph of \(y=f(x)\), indicating the exact \(x\)-intercept and the approximate location of \(P\).[3]
Solution:
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Answer:(a) quotient rule gives \(-\dfrac{1+2\ln x}{x^3}\) (b) \(P\left(\mathrm{e}^{-\frac12},\ \dfrac{\mathrm{e}}{2}\right)\) (c) \(f''\left(\mathrm{e}^{-\frac12}\right)=-2\mathrm{e}^2<0\), so a maximum (d) \(x>\dfrac{1}{\mathrm{e}}\) (e) curve rising from \(-\infty\) near the \(y\)-axis, crossing the \(x\)-axis at \(\dfrac{1}{\mathrm{e}}\), maximum at \(P\), then decreasing towards the \(x\)-axis
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