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

In summary, the problem asks for the slope of the tangent line at x = 0, but you forgot to take the limit as x -> 0. So, you can solve the equation for x using the limit as x -> 0.
  • #1
Rusho
24
0
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
:grumpy:

I'm taking a break
 
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  • #2
Rusho said:
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
:grumpy:

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:
[tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

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

-Dan
 
  • #3
topsquark said:
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:
[tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

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.
Rusho said:
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:
[tex]\frac{2x ^ 2 + 3x}{x}[/tex] or (2x ^ 2 + 3x) / x
Now, you can simplify that expression a little bit further, right?
Can you go from here? :)
 
  • #4
Ok, so I just do this:

=2x^2 + 3x/x

=x(2x + 3) / x

=2x + 3
 
Last edited:
  • #5
As long as x is not 0!
 
  • #6
Well 0! = 1, so I don't think that would be a problem ;)
 
  • #7
ksinclair13 said:
Well 0! = 1, so I don't think that would be a problem ;)
Err, I don't understand, why is factorial involved in here? :confused:
 
  • #8
It was joke, son, a joke! (The ;) was a give-away)

ksinclair13, you can get the actual :wink: by typing ": w i n k :" without the spaces or by going to "advanced" and clicking on the icon.
 
  • #9
Okay, thank you. :wink:

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

What is the rate of change from 0 to x in the function 2x^2 + 3x?

The rate of change from 0 to x in a function is also known as the slope or the derivative of the function at a particular point. In the case of 2x^2 + 3x, the rate of change from 0 to x is simply the coefficient of x, which is 3.

How do you calculate the rate of change from 0 to x in a quadratic function?

In a quadratic function, the rate of change from 0 to x can be calculated by finding the derivative of the function. For the function 2x^2 + 3x, the derivative is 4x + 3. Plugging in the value of x at the desired point will give you the rate of change from 0 to x.

What does the rate of change from 0 to x tell us about a function?

The rate of change from 0 to x tells us the instantaneous rate at which the function is changing at a specific point. It gives us information about the steepness or slope of the function at that point.

How can the rate of change from 0 to x be used in real-world applications?

The rate of change from 0 to x can be used in various real-world applications, such as calculating the velocity of a moving object or the rate of growth of a population. In these cases, the function would represent the distance or size at a given time, and the rate of change from 0 to x would give us information about how that distance or size is changing at a specific time.

What does a positive or negative rate of change from 0 to x indicate about a function?

A positive rate of change from 0 to x indicates that the function is increasing at that point, while a negative rate of change indicates that the function is decreasing. A rate of change of 0 indicates that the function is not changing at that point.

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