F'(x)= y ' = (4x² + 1/x )' = (4x²)' +(1/x) = 8x -1/x² ;
f'(x) =0 (критическая точка);
8x -1/x² = 0 ⇒ x³ = 1/8 ⇔ x₁= 1/2 .
f(a) = f(1/4) =4*(1/4)² +1/(1/4) = 1/4 +4 =4 ,25 ;
f(b) =4*1 +1/1 = 5;
f(x₁) =f(1/2) =4*(1/2)² + 1/(1/2) =1+2 =3.
miny =3;
maxy =5.
miny + maxy =3+5 =8.
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ответ : 8.