Correct proof (in stewart calculus) of fermat's theoram ?

[ManishR]

the proof of http://en.wikipedia.org/wiki/Fermat%27s_theorem_%28stationary_points%29" from stewart calculus is in the pdf.

In short, the author has done something like this,
consider,
$$m\leq0$$

$$\Rightarrow\frac{m}{n}\leq0$$

so

$$\frac{m}{n}\leq0$$ if $$n>0$$ .....[1]

and

$$\frac{m}{n}\geq0$$ if $$n<0$$ .....[2]

from inqualties [1] & [2]

$$\frac{m}{n}=0$$

$$\Rightarrow m=0$$

which is incorrect because in
n of [1] =/= n of [2]

correct proof


Fermat Theoram Statement:
If

$$f(c)\geq f(c+k)$$ OR $$f(c)\leq f(c+k)$$ where $$a-c\leq k\leq b-c$$

Then

$$\left[\frac{df(x)}{dx}\right]_{x=c}=0$$

Proof:

$$\left[\frac{df(x)}{dx}\right]_{x=c}=\frac{df(c)}{dx}$$

$$\Rightarrow\left[\frac{df(x)}{dx}\right]_{x=c}=\frac{dM}{dx}$$ where $$M=f(c)$$

$$\Rightarrow\left[\frac{df(x)}{dx}\right]_{x=c}=lim_{h\rightarrow0}$$$$\frac{M(x+h)-M(x)}{h}$$

$$\Rightarrow\left[\frac{df(x)}{dx}\right]_{x=c}=lim_{h\rightarrow0}$$$$\frac{0}{h}$$ because $$M(x+h)=M(x)$$

$$\Rightarrow\left[\frac{df(x)}{dx}\right]_{x=c}=0$$

so question is am i right ?

 

[Office_Shredder]

Your proof is incorrect. $$\left[ \frac{df}{dx}\right]_{x=c}$$ is NOT the same thing as differentiating f(c). The latter will give zero no matter what point c is, because f(c) is just a constant. What you calculated is the derivative of f(c) on the right side of each of those equalities

The idea is that the derivative must be equal to the one sided limit as h goes to zero of $$\frac{f(c+h)-f(c)}{h}$$. For very small values of h, this is going to be close to some number L.

If c is a maximum, then when h is positive the fraction that we have is positive, so it must be that L is larger than or equal to zero (because L IS the limit). On the other hand, if h is negative the fraction that we have is negative, so it must be that L is smaller than or equal to zero.

So we have $$0 \leq L \leq 0$$ which implies that L, which is f'(c), is zero.