Question

solve the problem....... ​

Answer 1

Step-by-step explanation:

function given, \bf{f(x)=\frac{logx}{x}}f(x)=

x

logx

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:

HOPE THE ATTACHMENT HELPS YOU