1 Answer | Add Yours
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.
We’ve answered 317,755 questions. We can answer yours, too.Ask a question