Answer:
(a)
The function f is continuous at [1,e] and differentiable at (1,e), therefore
the mean value theorem applies to the function.
(b)
= 1.71828
Step-by-step explanation:
(a)
The function f is continuous at [1,e] and differentiable at (1,e), therefore
the mean value theorem applies to the function.
(b)
You are looking for a point
such that
![\frac{1}{c} = \frac{\ln(15e)-\ln(15*1)}{e-1} = \frac{\ln(15e/15)}{e-1} = \frac{\ln(e)}{e-1} = \frac{1}{e-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bc%7D%20%3D%20%5Cfrac%7B%5Cln%2815e%29-%5Cln%2815%2A1%29%7D%7Be-1%7D%20%3D%20%5Cfrac%7B%5Cln%2815e%2F15%29%7D%7Be-1%7D%20%3D%20%5Cfrac%7B%5Cln%28e%29%7D%7Be-1%7D%20%3D%20%5Cfrac%7B1%7D%7Be-1%7D)
You have to solve for
and get that
= 1.71828
Answer:
With what?
Step-by-step explanation:
Answer:
12 _3/4
Step-by-step explanation:
3/4 + 1/2 + 2 = 3_1/4
16 cups in a gallon
16-3_1/4
=12_3/4
By the binomial theorem,
![\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%28x%5E2-%5Cfrac1x%5Cright%29%5E6%20%3D%20%5Csum_%7Bk%3D0%7D%5E6%20%5Cbinom%206k%20%28x%5E2%29%5E%7B6-k%7D%20%5Cleft%28-%5Cfrac1x%5Cright%29%5Ek%20%3D%20%5Csum_%7Bk%3D0%7D%5E6%20%5Cbinom%206k%20%28-1%29%5Ek%20x%5E%7B12-3k%7D)
I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that
12 - 3k = 0 ===> 3k = 12 ===> k = 4
which corresponds to the constant coefficient
![\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}](https://tex.z-dn.net/?f=%5Cdbinom%2064%20%28-1%29%5E4%20%3D%20%5Cdfrac%7B6%21%7D%7B4%21%286-4%29%21%7D%20%3D%20%5Cboxed%7B15%7D)