Differentiation of functions I was studying for my examination and came across this question:
Consider $f(x) = x \ln x$,
a.) For what values of $x$ is $f(x)$ defined? 
b.) Show that the global minimum value of $f(x)$ is $-1/e$.
I found the answer for a to be $x > 0$, however b is the one I am struggling with.
I have no idea where to start, I tried taking the derivative but that didn't really seem to help. Thanks in advance. 
 A: Where is it defined?
The logarithm $\ln x$ is only defined when $x > 0$. So, strictly speaking $x \ln x$ can only be evaluated at points where $x>0$. However, interestingly, if you let $x>0$ and then let $x$ get closer and closer to zero, i.e. get smaller and smaller, the limit of $x \ln x$ as $x$ tends to zero is actually zero:
$$\lim_{x \to 0^+} x \ln x = 0$$
Maxima and Minima
To find maxima and minima of any (differentiable) function, you can look at when the derivative vanishes. In your case $\mathrm{f}(x) = x\ln x$, which is a product and you may use the product rule to differentiate. The product rule says that $(uv)' = u'v + uv'$ and hence:
$$\mathrm{f}'(x) = 1\cdot \ln x + x\cdot \frac{1}{x} \equiv \ln x +1$$
It follows that $\mathrm{f}'(x)=0 \iff \ln x + 1 = 0 \iff \ln x =-1 \iff x = \mathrm{e}^{-1}=\frac{1}{\mathrm{e}}$. To find the value that $x \ln x$ takes when $x=\mathrm{e}^{-1}$ we simply substitute $x=\mathrm{e}^{-1}$:
$$\mathrm{f}(\mathrm{e}^{-1}) = \mathrm{e}^{-1} \cdot \ln(\mathrm{e}^{-1}) = \mathrm{e}^{-1} \cdot (-1) = -\mathrm{e}^{-1}=-\frac{1}{\mathrm{e}}$$
Proving its a minimum
To show that it is a minimum, we simply need to show that the second derivative satisfies $\mathrm{f}''(\mathrm{e}^{-1}) > 0$. Calculating the second derivative (by differentiating $\mathrm{f}'(x)=\ln x +1$) gives:
$$\mathrm{f}''(x) = \frac{1}{x} + 0 \equiv \frac{1}{x}$$
It follows that $\mathrm{f}''(\mathrm{e}^{-1}) = 1/\mathrm{e}^{-1} = \mathrm{e} > 0$.
A: So you took the derivative and obtained
$$
df/dx=  \ln x + 1 
$$
(presumably).
You know that the minimum will occur when $df/dx = 0$ 
And so we have $\ln x + 1 = 0$
This will give us a value for $x$, and substituting back into the original equation will allow us to evaluate $f(x)$ at its minimum value.
A: The function $f(x):=x\>\log x$ is defined for $x>0$, i.e., on ${\mathbb R}_{>0}$. To begin with we note the following facts:
$$\lim_{x\to0+} f(x)=0,\quad \lim_{x\to\infty} f(x)=\infty, \quad f\biggl({1\over2}\biggr)=-{1\over2}\log 2<0\ .$$
It follows that $f$ takes a global minimum $\mu<0$ on ${\mathbb R}_{>0}$, and this minimum is assumed at an interior point $\xi$ of the domain. In this case necessarily $f'(\xi)=0$. From
$f'(x)=1+\log x$ it then follows that $\ \xi={\displaystyle{1\over e}}$, so that $\mu$ becomes $$\mu=f\biggl({1\over e}\biggr)=-{1\over e}\ .$$
It has not been necessary (or even helpful) to compute second derivatives here.
