Fix: solution de l'exercice 3

1ST/03_Nombre_derive_et_tangente/exercises.tex

@@ -95,7 +95,7 @@
                 f(-3) = -2\times (-3) + 10 = 16
             \]
             \[
-                \frac{f(5) - f(-3)}{5 - (-3)} = \frac{2 - 16}{5 + 3} = \frac{-14}{8} = \frac{-7}{4}
+                \frac{f(4) - f(-3)}{4 - (-3)} = \frac{2 - 16}{4 + 3} = \frac{-14}{7} = -2
             \]
         \item
             \[
@@ -104,7 +104,7 @@
                 f(5) = 5^2 + 5 + 1 = 31
             \]
             \[
-                \frac{f(10) - f(5)}{10 - 5} = \frac{111 - 31}{10 - 5} = \frac{142}{5} = 28.4
+                \frac{f(10) - f(5)}{10 - 5} = \frac{111 - 31}{10 - 5} = \frac{80}{5} = 16
             \]
         \item
             \[