Someone please help me with this question? Thanks if you do!
1 answer:
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Answer:
<u>Step-by-step explanation:</u>
There are 3 conditions that must be satisfied:
- f(x) is continuous on the given interval
- f(x) is differentiable
- f(a) = f(b)
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
2. f'(x) = 3x² - 2x - 6 so the function is differentiable
3. f(0) = 0³ - 0² - 6(0) + 2 = 2
f(3) = 3³ - 3² - 6(3) + 2 = 2
f(0) = f(3)
f'(x) = 3x² - 2x - 6 = 0
This is not factorable so you need to use the quadratic formula:
Only one of these values (1.8) is between 0 and 3.