# Prove that ln2 is the smallest value of the function f(x)=ln[1+square root(1+x^2)].

### 2 Answers | Add Yours

To find the smallest value of the function we find the derivative and equate it to zero.

f(x) = ln [ 1 + sqrt (1 + x^2)]

f'(x) = [((1/2)/sqrt (1 + x^2))*2x]/[ 1 + sqrt (1 + x^2)]

equating this to zero

=> [((1/2)/sqrt (1 + x^2))*2x]/[ 1 + sqrt (1 + x^2)] = 0

=> [((1/2)/sqrt (1 + x^2))*2x] = 0

=> x = 0

Now f(0) = ln [ 1 + sqrt (1 + 0^2)]

=> ln ( 1 + sqrt 1)

=> ln 2

Also, any value of x does not give f(x) > ln 2.

**Therefore we can conclude that ln 2 is the smallest value of the function f(x) = ln [ 1 + sqrt (1 + x^2)].**

To determine the extreme value of a function we'll have to determine the critical point of the function. The critical point of the function is the root of the first derivative of the function.

We'll have to calculate the first derivative of the given function. Since f(x) is a composed function, we'll apply the chain rule.

f'(x) = ln[1+sqrt(1+x^2)]'

f'(x) = [2x/2sqrt(1+x^2)]/[1+sqrt(1+x^2)]

We'll simplify and we'll get:

f'(x) = x/sqrt(1+x^2)*[1+sqrt(1+x^2)]

We'll put f'(x) = 0

Since sqrt(1+x^2)>0, only the numerator could be zero.

x = 0

The critical point of the function is x = 0.

**The minimumvalue of the function is:**

f(0) = ln[1+sqrt(1+0^2)]

f(0) = ln(1+1)

**f(0) = ln2 q.e.d.**