Fermat's Theorem Stationary Points Proof

Fermat’s Theorem

• f’(c) exists
• local max/min at f(c) Then, $f^{\prime }(c)=0$

Proof

1. Assume $f^{\prime }(c)$ exists
   1. Suppose $f(c)$ is a local minima
      1. Then, by definition of local minima, $\exists \delta >0$ s.t $0<|x-c|<\delta ⟹f(x)\ge f(c)$
      2. Let $\delta$ be arbitrary
      3. Suppose $0<|x-c|<\delta$
      4. Then $f(x)\ge f(c)$
      5. This means $f(x)-f(c)\ge 0$
      6. Since $f^{\prime }(c)$ exists, then this means that $f(c)$ is continuous. Thus, ${\lim }_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}={\lim }_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}$
      7. Then, consider ${\lim }_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}$
         1. ${\lim }_{x\to c^{-}}(x-c)<0$ as $x\to c^{-}$
         2. Thus, ${\lim }_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}\le 0$
      8. Then, consider ${\lim }_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}$
         1. ${\lim }_{x\to c^{+}}(x-c)>0$ as $x\to c^{+}$
         2. Thus, ${\lim }_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}\ge 0$
      9. $0\ge {\lim }_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}={\lim }_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}\ge 0$
      10. Thus, ${\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}=0$
      11. Thus, $f^{\prime }(c)=0$
   2. Suppose $f(c)$ is a local maxima
      1. Then, by definition of local maxima, $\exists \delta >0$ s.t $0<|x-c|<\delta ⟹f(x)\le f(c)$
      2. Let $\delta$ be arbitrary
      3. Suppose $0<|x-c|<\delta$
      4. Then $f(x)\le f(c)$
      5. This means $f(x)-f(c)\le 0$
      6. Since $f^{\prime }(c)$ exists, then this means that $f(c)$ is continuous. Thus, ${\lim }_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}={\lim }_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}$
      7. Then, consider ${\lim }_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}$
         1. ${\lim }_{x\to c^{-}}(x-c)<0$ as $x\to c^{-}$
         2. Thus, ${\lim }_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}\ge 0$
      8. Then, consider ${\lim }_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}$
         1. ${\lim }_{x\to c^{+}}(x-c)>0$ as $x\to c^{+}$
         2. Thus, ${\lim }_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}\le 0$
      9. $0\le {\lim }_{x\to c^{-}}\frac{f(x)-f(c)}{x-c}={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}={\lim }_{x\to c^{+}}\frac{f(x)-f(c)}{x-c}\le 0$
      10. Thus, ${\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}=0$
      11. Thus, $f^{\prime }(c)=0$