Proof of Constant Rule: f'(x)=0

[TheKracken]

Homework Statement

For the Constant function f(x)=c prove that f'(x)= 0

Homework Equations

Alright so the relevant equation would be the Difference quotient or definition of a derivative.

The Attempt at a Solution

This is what I said

I restated the equation as this
f(x)=c >>>> f(c)= c(x^0)

Then I imputed into definition of derivative

(c(x+h)^0 - c(x^0))/ h

Then I solved it to be

(c-c)/h = 0/h
and then set h to 0 and then you get 0/0 which is non determinant form. I am not sure if this "proves" the statements to be true or if since it is 0/0 that doesn't show anything.

 

[LCKurtz]

Homework Statement

For the Constant function f(x)=c prove that f'(x)= 0

Homework Equations

Alright so the relevant equation would be the Difference quotient or definition of a derivative.

The Attempt at a Solution

This is what I said

I restated the equation as this
f(x)=c >>>> f(c)= c(x^0)

Then I imputed into definition of derivative

(c(x+h)^0 - c(x^0))/ h

You don't need an artificial $x^0$ factor. It is just $c-c$.
Then I solved it to be

(c-c)/h = 0/h = 0
and then set h to 0 and then you get 0/0 which is non determinant form. I am not sure if this "proves" the statements to be true or if since it is 0/0 that doesn't show anything.[/QUOTE]

That comes out zero before you let $h\to 0$ so the limit is zero. It is not indeterminante 0/0. So you're done.

 

[TheKracken]

So by including the artificial x^0 I actually made the proof incorrect? I am curious what is the most proper way to represent this proof? As I am going to be a math major, I don't really know how they would prefer this to be represented.

 

[LCKurtz]

So by including the artificial x^0 I actually made the proof incorrect? I am curious what is the most proper way to represent this proof? As I am going to be a math major, I don't really know how they would prefer this to be represented.

I would write it like this. Given $f(x)=c$ then$$
f'(0) = \lim_{h\to 0}\frac {f(0+h)-f(0)}{h} = \lim_{h\to 0}\frac {c-c}{h}=
\lim_{h\to 0}\frac {0}{h}=\lim_{h\to 0} 0=0$$to be explicit and correct.

 

[TheKracken]

did you mean f'(c) ?

 

[LCKurtz]

Woops, I was thinking you wanted f'(0). Just put $x$ instead of $0$ in that difference quotient in the first two expressions. It's the same arithmetic.

 

[verty]

Also you need to understand or be able to see that the limit is 0. I mean this expression: $lim_{h \to 0} {0 \over h}$ is 0 for every h, it is a horizontal line heading toward the origin, so of course the limit is 0.