# question about mean value

• December 5th 2009, 02:06 PM
dangerous1
question about mean value
f(x) = x^3 - 5x^2 - 3x
find the mean value theorem to find all numbers c in the interval (1,3)

i got the answer 5 +- (sqrt 16) / 3
am i correct
• December 5th 2009, 03:39 PM
tonio
Quote:

Originally Posted by dangerous1
f(x) = x^3 - 5x^2 - 3x
find the mean value theorem to find all numbers c in the interval (1,3)

i got the answer 5 +- (sqrt 16) / 3
am i correct

What is "all numbers c in the interval (1,3)"??
And no, you cannot be correct, whatever c is, since both numbers $5\pm \frac{\sqrt{16}}{3}=5\pm \frac{4}{3}$ are not in (1,3)...even if the numbers were $\frac{5\pm 4}{3}$ they still aren't in the interval (1,3).

Tonio
• December 5th 2009, 03:40 PM
skeeter
Quote:

Originally Posted by dangerous1
f(x) = x^3 - 5x^2 - 3x
find the mean value theorem to find all numbers c in the interval (1,3)

i got the answer 5 +- (sqrt 16) / 3
am i correct

$\frac{f(3) - f(1)}{3-1} = -10$

$f'(x) = 3x^2 - 10x - 3 = -10$

$3x^2 - 10x + 7 = 0$

factor and finish solving for $x$.