Question

If `f(x) = log (1 + x) + 1/(1 + x)`, show that f(x) attains its minimum value at x = 0.

Answer 1

Given, `f(x) = log (1 + x) + 1/(1 + x)`   ...(i)

Differentiate w.r.t. x

f'(x) = `1/(1 + x) - 1/(1 + x)^2`

⇒ f'(x) = `(1 + x - 1)/(1 + x)^2`

⇒ f'(x) = `x/(1 + x)^2`   ...(ii)

Take, f'(x) = 0

Now, `x/(1 + x)^2 = 0`

⇒ x = 0

Now, again differentiate w.r.t. x equation (ii)

f"(x) = `((1 + x)^2 d/dx x - x d/dx (1 + x)^2)/(1 + x)^4`

= `((1 + x)^2 - x xx 2(1 + x))/(1 + x)^4`

At x = 0

f"(0) = `(1 - 0)/1`

f"(0) = 1

f"(0) = 1 > 0

So, f(x) has a minimum value at x = 0.

Hence Proved.