# Given the function f(x)=2*square root x* ln (x). f:[,+infinite)->R, prove that f(x)>=-4/e.

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### 1 Answer

We'll prove that the right limit of the function, if x approaches to zero, x>0, is 0.

lim f(x) = 0, for x->0, x>0

lim f(x) = +infinite, for x-> +infinite

This fact demonstrates that the function is continuous over the range [0, +infinite).

Since the function is continuous, that means that it could be differentiated, with respect to x.

We'll apply product rule of differentiating.

f'(x) = lnx/sqrtx + 2sqrtx/x

We'll put f'(x) = 0

lnx/sqrtx + 2sqrtx/x = 0

We'll multiply all over by x*sqrtx

x*lnx + 2(sqrt x)^2 = 0

x*ln x + 2x = 0

We'll factorize by x:

x(ln x + 2) = 0

x = 0

ln x + 2 = 0

ln x = -2

x = e^-2

x = 1/e^2

This is a critical point for the given function. We'll calculate the ordinate of the extreme point, by substituting x into the expression of f(x).

f(1/e^2) = -4/e

We'll verify if the derivative is positive or negative, for values higher than 1/e^2. We'll put x = 1.

ln 1 + 2= 2 > 0

We'll verify if the derivative is positive or negative, for values smaller than 1/e^2. We'll put x = 1/e^3

ln e^-3 + 2 = -3+2 = -1 < 0

**The extreme point f(1/e^2) = -4/e is a minimum point so, f(x)>=-4/e.**