# Lowest value of simple function: $\sqrt{x}-\ln{x}$

$$y=\sqrt{x}-\ln{x}$$ as noted when graphed, the function has a lowest $y$ value when $x=4$. Can I have a rigorous proof of why this is the case?

• We will have to use calculus tools, precalculus is not enough. – André Nicolas Jul 19 '15 at 22:58
• @AndréNicolas Should I edit the tab there? – user253055 Jul 19 '15 at 22:59
• The derivative of $$\sqrt{x}- ln(x)$$ is $$\frac{1}{2\sqrt{x}- \frac{1}{x}= \frac{\sqrt{x}- 2}{2x}[/quote]. For any positive x, the denominator is positive so its sign is that of the numerator. For x< 4, [tex]\sqrt{x}$$ is less than 2 so the numerator is negative, That means the function value is [b]decreasing[/b]. For x> 4, $$\sqrt{x}$$ is larger than 2 so the numerator is positive. That means the function value is [b]increasing[/b]. – user247327 Jul 19 '15 at 23:03
• You edited, calculus is fine. – André Nicolas Jul 19 '15 at 23:05
• It doesn't make sense to have both calculus and pre-calculus as tags. It's either one or the other. – Zain Patel Jul 19 '15 at 23:08

I'm afraid that we're going to need to being in the big bad calculus for this particular proof. Let $f(x) = \sqrt{x} - \ln x$ then we have $$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{x} = \frac{\sqrt{x}-2}{2x}.$$ This means that the function has an extrema at $f'(x) = 0\iff \sqrt{x} = 2 \implies x = 4$.

Since we have $f(1) = 1 > 2 - \ln 4 = f(4)$ and $f(16) = 4 - \ln 16 > 2 - \ln 4 = f(4)$ then the point must be a minimum.

• I believe you're missing a square root sign in your $f'(x)$ expression – Brenton Jul 19 '15 at 23:10
• @Brenton, Woopsie daisy, thanks! :-) – Zain Patel Jul 19 '15 at 23:11
• @molarmass... I must be very tired. Wow. – Zain Patel Jul 19 '15 at 23:16

Let $f$ be the function defined by $x\mapsto\sqrt{x}-\ln x$. Since

$$f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x}=\frac{\sqrt{x}-2}{2x}$$

vanishes only for $x=4$, and $f''(4)=\frac{\sqrt{x}-2(\sqrt{x}-2)}{4x^2}_{|x=4}=\frac{1}{8}>0$ it follows that $f$ reaches its minimum value at $x=4$.