diff --git a/education/math/MATH1210 (calc 1)/Max and Min.md b/education/math/MATH1210 (calc 1)/Max and Min.md
index 1c834c4..3e0ae98 100644
--- a/education/math/MATH1210 (calc 1)/Max and Min.md	
+++ b/education/math/MATH1210 (calc 1)/Max and Min.md	
@@ -16,9 +16,22 @@ A local max/min is a peak or trough at any point along the graph.
 # Extreme Value Theorem
 If $f$ is a continuous function in a closed interval $[a, b]$, then $f$ achieves both an absolute maximum and an absolute minimum in $[a, b]$. Furthermore, the absolute extrema occur at $a$ or at $b$ or at a critical number between $a$ and $b$.
 
-## Example
+## Examples
 > Find the absolute maximum and absolute minimum of the function $f(x) = x^2 -3x + 2$ on the closed interval $[0, 2]$:
+
 1. $x=0$ and $x=2$ are both critical numbers because they are endpoints. Endpoints are *always* critical numbers because $\dfrac{d}{dx}$ is undefined.
 2. $\dfrac{d}{dx} x^2 -3x + 2 = 2x -3$
 3. Setting the derivative to zero, $0 = 2x - 3$
-4. Solving for x, we get $x = \dfrac{3}{2}$. Three halves is a critical number because $
\ No newline at end of file
+4. Solving for x, we get $x = \dfrac{3}{2}$. Three halves is a critical number because $f'(\dfrac{3}{2})$ is $0$. 
+5. Now check the outputs for all critical numbers ($f(x)$ at $x = \{0, 2, \dfrac{3}{2}\}$)
+6. $f(0) = 0^2 -3(0) + 2 = 2$
+7. $f(2) = 2^2 - 3(2) + 2) = 0$
+8. $f(\dfrac{3}{2}) = (\dfrac{3}{2})^2 - 3(\dfrac{3}{2}) + 2 = -\dfrac{1}{4}$
+9. The minimum is the lowest of the three, so it's $-\dfrac{1}{4}$ and it occurs at $x = \dfrac{3}{2}$ 
+10. The maximum is the highest of the three, so it's $2$ at $x = 0$.
+
+
+> 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.