Calculus: use algebra to evaluate the limit at x = 2

[sareevan]

It can be shown that if f(x) = 2x^2-3x+4, then f'(2) = lim [2(2+h)^2-3(2+h)+4]-[2(2)^2-3(2)+4]/h

Use algebra to evaluate this limit.

 

[galactus]

\(\displaystyle \L\\\lim_{h\to\0}\frac{(2(2+h)^{2}-3(2+h)+4)-(2(2)^{2}-3(2)+4)}{h}\)

Go ahead, see what you come up with. We'll be here to help you along.

The algebra ain't bad.

Take the derivative of your function the easy way, sub in x=2 and see what you get. If you perform the algebra correctly, you should get the same answer for your limit as h approaches 0.

 

[sareevan]

Calculus

I worked it out and came out with two answers. 5 and -1

 

[sareevan]

calculus

I need help figuring which was the right way.

 

[stapel]

I worked it out and came out with two answers. 5 and -1

Please reply showing all of your steps. (The tutors can't help you find your errors if they can't see your work.)

Thank you.

Eliz.

 

[sareevan]

calculus

(8+8h+2h^2) - (6-3h)-8+6-4/h = 8+8h+2h^2-6-3h-6 = -4+5h+2h^2 =

-4+2h^2 = √-4+2h^2 = -4+1 = -3

 

[stapel]

Re: calculus

(8+8h+2h^2) - (6-3h)-8+6-4/h = 8+8h+2h^2-6-3h-6 = -4+5h+2h^2 =

-4+2h^2 = √-4+2h^2 = -4+1 = -3

What happened to the "divided by h"? Where did the square roots come from?

Please show your work clearly, step-by-step. For instance, you have:

. . . . .[2(2 + h)² - 3(2 + h) + 4]

How did you square out the (2 + h)²? What did you get when you took the 2 through the square? What did you get when you took the 3 through the (2 + h)? What did you get when you simplified?

Then you have:

. . . . .[2(2)² - 3(2) + 4]

What did you get when you simplified this.

What did you get when you combined the two simplified expressions?

What did you get when you divided by "h"? Were you able to factor an "h" out of the numerator and then cancel off?

Thank you.

Eliz.