[m]F`(x)=(\sqrt{4x-5})`=\frac{1}{2\sqrt{4x-5}}\cdot (4x-5)`=\frac{1}{2\sqrt{4x-5}}\cdot (4)=\frac{2}{\sqrt{4x-5}}=f(x)[/m]
[m]F`(x)=f(x)[/m] для любого [m] x ∈ [\frac{5}{4};+ ∞ )[/m]
О т в е т. Да
2)
[m]F`(x)=(\sqrt{5-4x})`=\frac{1}{2\sqrt{5-4x}}\cdot (5-4x)`=\frac{1}{2\sqrt{5-4x}}\cdot (-4)=-\frac{2}{\sqrt{4x-5}}=f(x)[/m]
[m]F`(x)=f(x)[/m] для любого [m] x ∈ (- ∞; \frac{5}{4}][/m]
О т в е т. Да