Difference Quotient, fifth power

**Shonard** · December 5th 2009, 03:25 PM

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?

**VonNemo19** · December 5th 2009, 04:07 PM

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?

**Prove It** · December 5th 2009, 04:10 PM

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 $\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$ ?

**Shonard** · December 5th 2009, 04:11 PM

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

**Pulock2009** · December 5th 2009, 04:14 PM

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).

**Prove It** · December 5th 2009, 04:14 PM

Read my post.

**VonNemo19** · December 5th 2009, 04:15 PM

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...

**Shonard** · December 5th 2009, 04:25 PM

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?

**Prove It** · December 5th 2009, 04:35 PM

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

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

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

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

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

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

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

Now let $h \to 0$ and we find

$f'(x) = 10x^4 - 6x$ .

**Shonard** · December 5th 2009, 04:38 PM

Ah, I see my mistake thank you.

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