# Difference Quotient, fifth power

• Dec 5th 2009, 03:25 PM
Shonard
Difference Quotient, fifth power
Hello, I have a question regarding difference quotients. I haven't been able to find a single example of a difference quotient to an exponent higher than 3, so I was hoping someone here could give me a hand.

The exact wording is:

Let f(x) = 2x^5-3x^2+1

What is the difference quotient f' of f?

I solved it to the best of my knowledge, but the answer is rather long which causes me to doubt it. Any thoughts?
• Dec 5th 2009, 04:07 PM
VonNemo19
Quote:

Originally Posted by Shonard
Hello, I have a question regarding difference quotients. I haven't been able to find a single example of a difference quotient to an exponent higher than 3, so I was hoping someone here could give me a hand.

The exact wording is:

Let f(x) = 2x^5-3x^2+1

What is the difference quotient f' of f?

I solved it to the best of my knowledge, but the answer is rather long which causes me to doubt it. Any thoughts?

The difference quotient can be defined (geometrically) as the slope of a secant line through the graph of a function $f(x)$. In other words

The difference quotient: $\frac{\Delta{y}}{\Delta{x}}$.

But your problem is asking you to find $f'(x)=\lim_{\Delta{x}\to0}\frac{\Delta{y}}{\Delta{ x}}$.

Do you know how to do this?
• Dec 5th 2009, 04:10 PM
Prove It
Quote:

Originally Posted by Shonard
Hello, I have a question regarding difference quotients. I haven't been able to find a single example of a difference quotient to an exponent higher than 3, so I was hoping someone here could give me a hand.

The exact wording is:

Let f(x) = 2x^5-3x^2+1

What is the difference quotient f' of f?

I solved it to the best of my knowledge, but the answer is rather long which causes me to doubt it. Any thoughts?

Do you know the Binomial Theorem? It will help you to expand $(x + h)^5$.

$(a + b)^n = \sum_{r = 0}^n {n\choose{r}} a^{n - r}b^r$, provided $n$ is a positive integer.

So that would mean $(x + h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5$.

Do you think now you can evaluate $\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$?
• Dec 5th 2009, 04:11 PM
Shonard
I know how to solve the difference quotient through the formula..

f(x+h)-f(x)/h

the fifth power is making it tricky though
• Dec 5th 2009, 04:14 PM
Pulock2009
you are correct. the answer would be rather lengthy.
f(x)=2x^5-3x^2+1
f(x+h)=2(x+h)^5-3(x+h)^2+1

f'(x)=lim h->0 f(x+h)-f(x)/h

u will have to expand (x+h)^5 using binomial theorem which makes the problem lengthy. at last h would get cancelled. and u will get f'(x)=10x^4-6x
which is nothing but the differentiation of f(x).
• Dec 5th 2009, 04:14 PM
Prove It
• Dec 5th 2009, 04:15 PM
VonNemo19
Quote:

Originally Posted by Shonard
I know how to solve the difference quotient through the formula..

f(x+h)-f(x)/h

the fifth power is making it tricky though

As prove it mentioned, if you insist upon differentiating this with the definition, then the binomial theorem will help you. Or, perhaps, Pascal's triangle...
• Dec 5th 2009, 04:25 PM
Shonard
Quote:

Originally Posted by Prove It
Do you know the Binomial Theorem? It will help you to expand $(x + h)^5$.

$(a + b)^n = \sum_{r = 0}^n {n\choose{r}} a^{n - r}b^r$, provided $n$ is a positive integer.

So that would mean $(x + h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5$.

Do you think now you can evaluate $\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$?

Ok I attempted to evaluate using the expansion of (x+h)^5 and got:

$10x^4 + 20x^3h + 20x^2h^2 + 10xh^3 + 2h^4 - 2x - h$

does this seem correct?
• Dec 5th 2009, 04:35 PM
Prove It
$f(x) = 2x^5 - 3x^2 + 1$.

$f(x + h) = 2(x + h)^5 - 3(x + h)^2 + 1$

$= 2(x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5) - 3(x^2 + 2xh + h^2) + 1$

$= 2x^5 + 10x^4h + 20x^3h^2 + 20x^2h^3 + 10xh^4 + 2h^5 - 3x^2 - 6xh - 3h^2 + 1$.

$f(x + h) - f(x) = 2x^5 + 10x^4h + 20x^3h^2 + 20x^2h^3 + 10xh^4 + 2h^5 - 3x^2 - 6xh - 3h^2 + 1$
$- (2x^5 - 3x^2 + 1)$

$= 10x^4h + 20x^3h^2 + 20x^2h^3 + 10xh^4 + 2h^5 - 6xh - 3h^2$.

$\frac{f(x + h) - f(x)}{h} = 10x^4 + 20x^3h + 20x^2h^2 + 10xh^3 + 2h^4 - 6x - 3h$.

Now let $h \to 0$ and we find

$f'(x) = 10x^4 - 6x$.
• Dec 5th 2009, 04:38 PM
Shonard
Ah, I see my mistake thank you.