Answer

A) The problem of founding the extrema a given function is always equivalent with finding the zeros of the first derivative of the function.

$f(x) = 10x – x^2$

$f'(x) = 10 – 2x$, which means that the extrema of the function is at the point where $f'(x) = 0$,

$x = 10/2 = 5$

because $x – \alpha <=0$, where $\alpha$ is real, the function has an extrema for all $\alpha$ values: $\alpha >=5$. Otherwise there is no extrema. in the case $\alpha >=5$ the extrema is

$f(5) = 10*5 -5*5 = 25$

B) In this case $f(x) = ln(x)$, which means that $f'(x)= 1/x$.

One can simply observe that this expression in zero only for the values of x equal to positive or negative infinite.

(x = +,-infinite).

Because by its definition the function g(x) is zero only on the interval x, [0,1] (and in rest always positive) this means that there are no values of x to meet the required criteria for the derivative of the function f to be zero. In turn, it means that the function f(x) has no extrema.