First, make sure the mean value theorem applies for the given <em>f</em> . Its domain is <em>x</em> + 1 > 0, or <em>x </em>> -1, and it's continuous and differentiable on its domain, so all is well.
By the MVT, we have for some <em>c</em> in the open interval (1, 4),
<em>f '</em> (<em>c</em>) = (<em>f</em> (4) - <em>f</em> (1)) / (4 - 1)
<em>f</em> (<em>x</em>) = 5 cos²(<em>x</em>/2) + ln(<em>x</em> + 1) - 3
→ <em>f</em> (4) = 5 cos²(2) + ln(5) - 3
→ <em>f</em> (1) = 5 cos²(1/2) + ln(2) - 3
→ <em>f '</em> (<em>c</em>) = (5 cos²(2) + ln(5) - 3 - 5 cos²(1/2) - ln(2) + 3) / 3
-5 cos(<em>c</em>/2) sin(<em>c</em>/2) + 1/(<em>c</em> + 1) = (5 (cos²(2) - cos²(1/2)) + ln(5/2)) / 3
-15 sin(<em>c</em>) + 6/(<em>c</em> + 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 < <em>c</em> < 4, there are two solutions <em>c</em> ≈ 1.0525 and <em>c</em> ≈ 2.217.