So sánh \(f\left( { - \frac{1}{2}} \right)\) và \...

B

- Hướng dẫn giải

Phương pháp giải:

Thay \(x =  - \frac{1}{2}\) vào \(f\left( x \right)\) và \(g\left( x \right)\) để tính \(f\left( { - \frac{1}{2}} \right);\,\,\,g\left( { - \frac{1}{2}} \right)\) rồi so sánh. 

Giải chi tiết:

Ta có:

 \(\begin{array}{l}f\left( { - \frac{1}{2}} \right) = 5.{\left( { - \frac{1}{2}} \right)^5} - 6{\left( { - \frac{1}{2}} \right)^4} + 2{\left( { - \frac{1}{2}} \right)^3} + 4{\left( { - \frac{1}{2}} \right)^2} - 3.\left( { - \frac{1}{2}} \right) + 6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =  - \frac{5}{{32}} - 6.\frac{1}{{16}} + 2.\left( { - \frac{1}{9}} \right) + 4.\frac{1}{4} + \frac{3}{2} + 6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =  - \frac{5}{{32}} - \frac{3}{8} - \frac{2}{9} + 1 + \frac{3}{2} + 6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =  - \left( {\frac{5}{{32}} + \frac{3}{8} + \frac{2}{9} - \frac{3}{2}} \right) + 7\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{215}}{{288}} + 7\end{array}\)

\(\begin{array}{l}g\left( x \right)\, = \, - 5{\left( { - \frac{1}{2}} \right)^5} - 2{\left( { - \frac{1}{2}} \right)^4} + 2{\left( { - \frac{1}{2}} \right)^3} + 4{\left( { - \frac{1}{2}} \right)^2} - \left( { - \frac{1}{2}} \right) + 5\\\,\,\,\,\,\,\,\,\,\,\, = \,\frac{5}{{32}} - 2.\frac{1}{{16}} + 2.\left( { - \frac{1}{8}} \right) + 4.\frac{1}{4} + \frac{1}{2} + 5\\\,\,\,\,\,\,\,\,\,\,\, = \,\frac{5}{{32}} - \frac{1}{8} - \frac{2}{8} + 1 + \frac{1}{2} + 5\\\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{{32}} - \frac{3}{8} + \frac{1}{2} + 6\\\,\,\,\,\,\,\,\,\,\,\, = \frac{9}{{32}} + 6\\\,\,\,\,\,\,\,\,\,\,\, = \frac{{201}}{{32}} = 6,28125 < \frac{{225}}{{288}} + 7 = f\left( x \right)\end{array}\)

Vậy \(f\left( { - \frac{1}{2}} \right) > g\left( { - \frac{1}{2}} \right)\)

Chọn B