Answer: [tex]\bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}[/tex]
Step-by-step explanation:
There are 3 conditions that must be satisfied:
If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.
f(x) = x³ - x² - 6x + 2 [0, 3]
1. There are no restrictions on x so the function is continuous [tex]\checkmark[/tex]
2. f'(x) = 3x² - 2x - 6 so the function is differentiable [tex]\checkmark[/tex]
3. f(0) = 0³ - 0² - 6(0) + 2 = 2
f(3) = 3³ - 3² - 6(3) + 2 = 2
f(0) = f(3) [tex]\checkmark[/tex]
f'(x) = 3x² - 2x - 6 = 0
This is not factorable so you need to use the quadratic formula:
[tex]x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1[/tex]
Only one of these values (1.8) is between 0 and 3.
