diff --git a/education/math/MATH1210 (calc 1)/Max and Min.md b/education/math/MATH1210 (calc 1)/Max and Min.md
index 3e0ae98..dd94c2a 100644
--- a/education/math/MATH1210 (calc 1)/Max and Min.md	
+++ b/education/math/MATH1210 (calc 1)/Max and Min.md	
@@ -34,4 +34,11 @@ If $f$ is a continuous function in a closed interval $[a, b]$, then $f$ achieves
 > Find the absolute maximum and absolute minimum of the function $h(x) = x + 2cos(x)$ on the closed interval $[0, \pi]$.
 
  1. $x = 0$ and $x = \pi$ are both critical numbers because they are endpoints. Endpoints are critical because $\dfrac{d}{dx}$ is undefined.
- 2. 
+ 2. $\dfrac{d}{dx} x + 2\cos(x) = 1 - 2sin(x)$ 
+ 3. Setting that to zero, we get $0 = 1 - 2\sin(x)$
+ 4. $\sin(x) = \dfrac{1}{2}$ 
+ 5. In the interval $[0, \pi]$, $\sin(x)$ has a value of $\dfrac{1}{2}$ in two places: $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$. These are both critical numbers because they are points where $\dfrac{d}{dx}$ is zero.
+ 6. Now we plug these values into the original function:
+ 7. $h(0) = 0 + 2\cos 0 = 2$
+ 8. $h(\pi) = \pi + 2\cos(\pi) = \pi - 2$
+ 9. $h(\dfrac{pi}{6}) = \dfrac{\pi}{6} + 2\cos(\dfrac{pi}$