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]
