Difference quotient question

**Khaotical** · September 16th 2009, 09:11 PM

Find the difference quotient of f

f(x)=[7x^3]-6

Not really sure where to go with this one, maybe cause its late I'm not thinking correctly. Any help is greatly appreciated.

**Prove It** · September 16th 2009, 09:32 PM

Originally Posted by Khaotical

Find the difference quotient of f

f(x)=[7x^3]-6

Not really sure where to go with this one, maybe cause its late I'm not thinking correctly. Any help is greatly appreciated.

Is it

$f(x) = 7x^3 - 6$ ?

If so

$f(x + h) = 7(x + h)^2 - 6$

$= 7(x^2 + 2xh + h^2) - 6$

$= 7x^2 + 14xh + 7h^2 - 6$ .

Therefore

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

$= \lim_{h \to 0}\frac{7x^2 + 14xh + 7h^2 - 6 - (7x^2 - 6)}{h}$

$= \lim_{h \to 0}\frac{14xh + 7h^2}{h}$

$= \lim_{h \to 0}(14x + 7h)$

$= 14x$ .

**Matt Westwood** · September 16th 2009, 10:03 PM

Originally Posted by Prove It

Is it

$f(x) = 7x^3 - 6$ ?

If so

$f(x + h) = 7(x + h)^2 - 6$

$= 7(x^2 + 2xh + h^2) - 6$

$= 7x^2 + 14xh + 7h^2 - 6$ .

Therefore

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

$= \lim_{h \to 0}\frac{7x^2 + 14xh + 7h^2 - 6 - (7x^2 - 6)}{h}$

$= \lim_{h \to 0}\frac{14xh + 7h^2}{h}$

$= \lim_{h \to 0}(14x + 7h)$

$= 14x$ .

It's cubed, not squared, in the initial problem. But the technique is okay.

**Prove It** · September 17th 2009, 02:16 PM

Originally Posted by Matt Westwood

It's cubed, not squared, in the initial problem. But the technique is okay.

Oh yeah... oops...

$f(x) = 7x^3 - 6$ .

$f(x + h) = 7(x + h)^3 - 6$

$= 7(x^3 + 3x^2h + 3xh^2 + h^3) - 6$

$= 7x^3 + 21x^2h + 21xh^2 + 7h^3 - 6$ .

Now work out

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

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