Let $f(x) = 1 - \sqrt{x^2}$, where the square root is to be taken as positive. Then:

$A$ triangular garden is fenced on two sides,and the third side is a straight river bank. The two fenced sides have an equal length of $x$. What is the maximum area that can be enclosed by the garden?

The function $f(x) = \frac{x^2 - 2}{\sqrt{1 + x^2}}$:

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$,then

Let a function $f:[0,5] \rightarrow R$ be continuous. $f(1)=3$ and $F$ be defined as $F(x)=\int_{1}^{x} t^{2} g(t) dt$,where $g(t)=\int_{1}^{t} f(u) du$. Then for the function $F$,the point $x=1$ is

The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to