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Class 12th   Class 12th Maths

Miscellaneous Exercise on Chapter 6

Question 1

Show that the function given by

$$f(x)=\frac{\log x}{x}$$

has maximum at $x=e$.

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Solution

Given:

$$f(x)=\frac{\log x}{x},x>0$$

Differentiate using quotient rule:

$$f^{\mathrm{\prime }}(x)=\frac{(1\mathrm{/}x)\cdot x-(\log x)\cdot 1}{x^2}$$

Simplify numerator:

$$f^{\mathrm{\prime }}(x)=\frac{1-\log x}{x^2}$$

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To find critical points, set $f^{\mathrm{\prime }}(x)=0$

$$\frac{1-\log x}{x^2}=0$$

Since $x^2>0$ always,

$$1-\log x=0$$
$$\log x=1$$
$$x=e$$

So the critical point is $x=e$.

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Check whether it is maximum (Second Derivative Test)

Find $f^{\mathrm{\prime }\mathrm{\prime }}(x)$:

$$f^{\mathrm{\prime }}(x)=\frac{1-\log x}{x^2}$$

Differentiate again:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{-1\mathrm{/}x\cdot x^2-(1-\log x)\cdot 2x}{x^4}$$

Simplify numerator:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{-x-2x(1-\log x)}{x^4}$$
$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{-x-2x+2x\log x}{x^4}$$
$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{x(2\log x-3)}{x^4}$$
$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{2\log x-3}{x^3}$$

Now check at $x=e$:

$$f^{\mathrm{\prime }\mathrm{\prime }}(e)=\frac{2\cdot 1-3}{e^3}=\frac{-1}{e^3}<0$$

Since $f^{\mathrm{\prime }\mathrm{\prime }}(e)<0$, the function is concave down at $x=e$, therefore:

$$\boxed{{\displaystyle x=e\text{ is a point of maximum}}}$$

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Final Answer

$$\boxed{{\displaystyle \text{The function }f(x)=\frac{\log x}{x}\text{ has a maximum at }x=e\mathrm{.}}}$$