Question

Найти критические точки f(x)=5+12x-7x³

Answer 1

$f(x)=5+12x-7x^3$
$f'(x)=(5+12x-7x^3)'=12-21x^2$
$f'(x)=0$
$12-21x^2=0$
$21x^2=12$
$7x^2=4$
$x^2= \frac{4}{7}$
$x=б \sqrt{ \frac{4}{7} }$
$x=б \ \frac{2}{ \sqrt{7} } }$
$x=б \ \frac{2 \sqrt{7} }{7} }$