Stationary points of f(x)=x+√x

fx

The curves in the pic are f(x) and f'(x).

Question: The function is f(x)=x+√x. Can you find its min or max value? If you calculate f'(x) and put it to 0, you get x=1/4. But draw the curve and there is no such max/min there.

One suggestion:   Clearly f'(x) has two asymptotes at x=0 and y=1 and so can never be 0 for any real value of x. But why does differentiating lead to the result x=1/4? I agree that f(x) = x-√x has a clear minimum there, but it’s a different function. My answer was that since there’s an infinity at x=0 f'(x) is undefined, but I’m not convinced that’s the answer.

OK, so √x is defined as the positive value of x^(1/2).

Let’s see if we can differentiate it:

x = √x√x  where x, √x >= 0.

D(x)  = 1 = 2√x*D(x) => D(x) = 1/(2√x)

So, f'(x) = 1 + 1/(2√x)

if f'(x) =0, then √x = -1/2.  But √x is >= 0 by definition, so there is no solution.

Let’s try to be a little more systematic while staying within the bounds of high school maths.

We consider the function  g(x,c) = x + x^(1/2)

where x^(1/2) = (√x)*(-1)^c

and we choose c to be either 0 or 1.

So g'(x) = 1 + (1/2√x)*(-1)^c

Setting g'(x) = 0 =>

√x = (-1/2)*(-1)^c

But since  √x is positive, c must be 1 rather than 0 for this to have a solution.

So by accepting a solution here we force the original function to be g(x) = x – √x.

 

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