diff --git a/education/math/MATH1210 (calc 1)/Max and Min.md b/education/math/MATH1210 (calc 1)/Max and Min.md
index dd94c2a..01b79c7 100644
--- a/education/math/MATH1210 (calc 1)/Max and Min.md	
+++ b/education/math/MATH1210 (calc 1)/Max and Min.md	
@@ -40,5 +40,6 @@ If $f$ is a continuous function in a closed interval $[a, b]$, then $f$ achieves
  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. $h(\dfrac{pi}{6}) = \dfrac{\pi}{6} + 2\cos(\dfrac{pi}$
+ 8. $h(\pi) = \pi + 2\cos(\pi) = \pi - 2 \approx 1.14159$
+ 9. $h(\dfrac{\pi}{6}) = \dfrac{\pi}{6} + 2\cos(\dfrac{\pi}{6}) = \dfrac{\pi}{6} + 2(\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{6} + \sqrt{3} \approx 2.2556$ 
+ 10. $h(\dfrac{5\pi}{6}) = \dfrac{5\pi}{6} + 2\cos(\dfrac{5\pi}{6}) = \dfrac{5\pi}{6} - \sqrt{3} \approx 0.88594$