Let f be the function given by f (x) = 5 cos^2(x/2) + ln(x + 1) - 3. The derivative of f is given by f'(x)= -5 cos (x/2) sin(x/2) +
(1)/(x+1). What value of c satisfies the conclusion of the Mean Value
Theorem applied to f on the interval 1,4 ?

First, make sure the mean value theorem applies for the given f . Its domain is x + 1 > 0, or x > -1, and it's continuous and differentiable on its domain, so all is well.

By the MVT, we have for some c in the open interval (1, 4),

f ' (c) = (f (4) - f (1)) / (4 - 1)

f (x) = 5 cos²(x/2) + ln(x + 1) - 3

→   f (4) = 5 cos²(2) + ln(5) - 3

→   f (1) = 5 cos²(1/2) + ln(2) - 3

→   f ' (c) = (5 cos²(2) + ln(5) - 3 - 5 cos²(1/2) - ln(2) + 3) / 3

-5 cos(c/2) sin(c/2) + 1/(c + 1) = (5 (cos²(2) - cos²(1/2)) + ln(5/2)) / 3

-15 sin(c) + 6/(c + 1) = 10 (cos(4) - cos(1)) + 2 ln(5/2)

You'll need the help of a calculator to solve this. Over the interval 1 < c < 4, there are two solutions c ≈ 1.0525 and c ≈ 2.217.

5

Step-by-step explanation:

When you divide 1 by 6, you get 0.17. This means that the unknown number must have the value of 0.17 when divided. The only number that works with these requirements is 5, when divided by 30.