Question

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

Answer 1

function given, $\bf{f(x)=\frac{logx}{x}}$
differentiate f(x) with respect to x,
f'(x) = [x.1/x - logx]/x²
f'(x) = [1 - logx]/x²

now, f'(x) = 0
[1 - logx] = 0 => logx = 1
x = e

again, differentiate with respect to x,
f"(x) = {x²(-1/x) - (1 - logx).2x}/x⁴
= {-x - 2x(1 - logx)}/x⁴
now, put x = e
f"(e) = {- e - 2e(1 - loge)}/e⁴ = -1/e³ < 0
hence, it is clear that function f(x) has maximum at x = e

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.