So, f(x) = -3x^2 + 4 on interval [-1, 2]. what value of f ' (x) does the MVT assure us exists? at what point does it exist? for the first question, i got - 3 and for the second question, i got 1/2. Is this right? If not, how did you get your answer?
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f'(c) = f(b) - f(a) / (b-a)
f'(x) = -6x
-6x = -8 - (1) / (3)
-6x = -9 / 3
6x = 3
x = 1/2 (Where it exists)
What value it is f'(x) = secant line = -6x = -3.
Good job.
f'(x) = -6x
-6x = -8 - (1) / (3)
-6x = -9 / 3
6x = 3
x = 1/2 (Where it exists)
What value it is f'(x) = secant line = -6x = -3.
Good job.