Actually, those are the roots of your equation, but they have nothing to do with the problem.
Since f(-4) = f(3) = 0, the slope of the line segment joining (-4,0) and (3,0) is zero.
the MVT says there is at least one value c between -4 and 3 where f'(c) = 0.
So, since
f'(x) = 3x^2+10x-8
you want to find x where f'(x) = 0.
That is at x = -4 or x = 2/3
Not sure where you got the 7, but I suspect you tossed it in because that is the length of the interval. Take a second look at the MVT. It says that there is a c such that
(f(3) - f(-4))/7 = 0
Multiplying by 7 does not give you f'(x)=7.
i was doing the mean theorem for this problem:
(x+4)^2 (x-3) on [-4,3]
i got to:
3x^2 + 10x-8=7
someone helped me out and told me the answer was (-5 +/- sqrt(70))/(3)
but i don't know how to get to there, can someone please explain to me how to get to this point?
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