Rate of change from 0 to x 2x^2 + 3x

[Rusho]

2x^2 + 3x

I'm not really sure what to do with the "X"
Form: f(x) - f(c)/ x-c

-2(x)^2 +3(x) - (-2(0)^2 + 3 (0) / x-0

I get
= -2x^2 + 3x + 0 / x - 0

= -2x^2 + 3x / x

that doesn't seem right


I'm taking a break

 

[topsquark]

2x^2 + 3x

I'm not really sure what to do with the "X"
Form: f(x) - f(c)/ x-c

-2(x)^2 +3(x) - (-2(0)^2 + 3 (0) / x-0

I get
= -2x^2 + 3x + 0 / x - 0

= -2x^2 + 3x / x

that doesn't seem right


I'm taking a break

You forgot to take the limit as x -> 0.

Personally, I've never been too keen on that form for the derivative. I think it's more transparent to use:
$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

That might show you a bit better what happens to everything.

-Dan

 

[VietDao29]

You forgot to take the limit as x -> 0.

Personally, I've never been too keen on that form for the derivative. I think it's more transparent to use:
$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

That might show you a bit better what happens to everything.

-Dan

topsquark, you don't need to take the limit as x -> 0. The problem does not ask for the slope of the tangent line at x = 0. It just ask for the rate of change from 0 to x.

I get
= -2x^2 + 3x + 0 / x - 0

= -2x^2 + 3x / x

There should be no minus sign in front of the 2x², and you seem to be missing some parentheses. :)
It should read:
$$\frac{2x ^ 2 + 3x}{x}$$ or (2x ^ 2 + 3x) / x
Now, you can simplify that expression a little bit further, right?
Can you go from here? :)

 

[Rusho]

Ok, so I just do this:

=2x^2 + 3x/x

=x(2x + 3) / x

=2x + 3

 

[HallsofIvy]

As long as x is not 0!

 

[ksinclair13]

Well 0! = 1, so I don't think that would be a problem ;)

 

[VietDao29]

Well 0! = 1, so I don't think that would be a problem ;)

Err, I don't understand, why is factorial involved in here?

 

[HallsofIvy]

It was joke, son, a joke! (The ;) was a give-away)

ksinclair13, you can get the actual by typing ": w i n k :" without the spaces or by going to "advanced" and clicking on the icon.

 

[ksinclair13]

Okay, thank you.

I always use the quick reply, but now I know how to do it with that as well