Question

f'(x)/q'(x)= 0, если f(x)=x³ - 4x, q(x)= корень из x

Answer 1

$f(x)=x^3-4x\\q(x)= \sqrt{x} \\\\f`(x)=3x^2-4\; \; \; \; \; q`(x)= \frac{1}{2 \sqrt{x} } \\\\ \frac{f`(x)}{q`(x)}= \frac{3x^2-4}{ \frac{1}{2 \sqrt{x} } }=2 \sqrt{x}(3x^2-4) \\\\\frac{f`(x)}{q`(x)}=0\\\\2 \sqrt{x}(3x^2-4)=0\\\\ 1) \sqrt{x} =0\; \; =\ \textgreater \ x_1=0\\\\2) 3x^2-4=0\\\ 3x^2=4\\x^2= \frac{4}{3} \\x_{2,3}=б \sqrt{ \frac{4}{3} }$