Suriin ang IDNStudy.com para sa malinaw at detalyadong mga sagot. Makakuha ng mga sagot sa iyong mga tanong mula sa aming mga eksperto, handang magbigay ng mabilis at tiyak na solusyon.

Let f(x) = x² - 4x3 + 6x2 4x + 1. Prove that f (x) is always non-negative for all real numbers x.​

Sagot :

Step-by-step explanation:

[tex]f(x) = x^2 - 4x^3 + 6x^2 - 4x + 1[/tex]

Combine like terms:

[tex]f(x) = -4x^3 + 7x^2 - 4x + 1[/tex]

[tex]{f'(x) = \frac{d}{dx} (-4x^3 + 7x^2 - 4x + 1) }[/tex]

[tex]f'(x) = -12x^2 + 14x - 4[/tex]

Set the derivative to zero to find the critical points:

[tex]-12x^2 + 14x - 4 = 0[/tex]

[tex]( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):[/tex]

[tex]x = \frac{-14 \pm \sqrt{14^2 - 4(-12)(-4)}}{2(-12)}[/tex]

[tex]x = \frac{-14 \pm \sqrt{196 - 192}}{-24}[/tex]

[tex]x = \frac{-14 \pm \sqrt{4}}{-24}[/tex]

[tex]x = \frac{-14 \pm 2}{-24}[/tex]

[tex]x = \frac{-12}{-24} = \frac{1}{2}[/tex]

[tex]x = \frac{-16}{-24} = \frac{2}{3}[/tex]

Evaluate ( f(x) ) at these critical points:

[tex]{f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right)^3 + 6\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 1}[/tex]

[tex]= \frac{1}{4} - \frac{4}{8} + \frac{6}{4} - 2 + 1[/tex]

[tex]= \frac{1}{4} - \frac{1}{2} + \frac{3}{2} - 2 + 1[/tex]

[tex]= \frac{1}{4} - \frac{1}{2} + \frac{3}{2} - 2 + 1 [/tex]

[tex]= \frac{1}{4} - \frac{2}{4} + \frac{6}{4} - \frac{8}{4} + \frac{4}{4}[/tex]

[tex]= \frac{1 - 2 + 6 - 8 + 4}{4}[/tex]

[tex]= \frac{1}{4}[/tex]

[tex] \text{Similarly}, for \( x = \frac{2}{3} \):[/tex]

[tex]{f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right)^3 + 6\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 1}[/tex]

[tex]= \frac{4}{9} - 4\left(\frac{8}{27}\right) + 6\left(\frac{4}{9}\right) - \frac{8}{3} + 1[/tex]

[tex]= \frac{4}{9} - \frac{32}{27} + \frac{24}{9} - \frac{24}{9} + 1[/tex]

[tex]= \frac{4}{9} - \frac{32}{27} + \frac{24}{9} - \frac{24}{9} + 1[/tex]

[tex]= 1 - \frac{32}{27}[/tex]

[tex]= \frac{27}{27} - \frac{32}{27}[/tex]

[tex]= \frac{-5}{27}[/tex]