# Show that the function given by f (x) = log x/x has maximum at x = e

**Solution:**

Maxima and minima are known as the extrema of a function. Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

The given function is f (x) = log x / x

Therefore,

f' (x) = [x(1/x) log x]/x^{2}

= (1 - log x)/x^{2}

Now,

For maxima or minima

f' (x) = 0

⇒ 1 - log x = 0

⇒ log x = 1

As we know that log e = 1

⇒ log x = log e

⇒ x = e

Also,

f" (x) = [x^{2}(1/x) - (1 - log x)(2x)] / x^{4}

= (- x - 2x (1 - log x))/x^{4}

= (- 3 + 2 log x)/x^{4}

Now,

f" (e) = (- 3 + 2 log x)/e^{3}

= (- 3 + 2)/e^{3}

= - 1//e^{3} < 0

Since f" (e) < 0, we get maximum value.

Therefore, by the second derivative test,

f is the maximum at x = e

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 2

## Show that the function given by f (x) = log x/x has maximum at x = e.

**Summary:**

Hence by using the second derivative test, the function given by f (x) = log x/x has a maximum at x = e. Maxima and minima are the maximum or the minimum value of a function within the given set of ranges

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