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.
