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

Expert Answers

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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)].

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