Asked by eric
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?
(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?
Answers
Answered by
Steve
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.
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.
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