- #1

Chris_w

- 9

- 0

I am a little confused as to why they give you a point... how can I solve this?

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- Thread starter Chris_w
- Start date

- #1

Chris_w

- 9

- 0

I am a little confused as to why they give you a point... how can I solve this?

- #2

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

They just want f'(2) and are telling you that f(2) = -2.

- #3

Chris_w

- 9

- 0

ahh, tricky tricky :) Thanks hurkyl

- #4

FrostScYthe

- 80

- 0

f'(x) = -4(x-3)

f'(2) = -4(2-3)

f'(2) = 4

4 is the derivative of f(x) at point x=2 which is also THE SLOPE

and if you have a point (2,-2) and a SLOPE gives you an equation, which is the equation of the tangent line at that point. Tangent is same thing as Derivative except that in the derivative they are too lazy to write an equation for that line, they only put the slope

I hope that helps, don't hate me too much if i'm confusing you

- #5

hawaiidude

- 41

- 0

- #6

FrostScYthe

- 80

- 0

That just takes too long

- #7

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

hawaiidude postedor you can use teh "good" old lim h~~>0 f(x+deltax)-f(x)/ delta x...in your case f(x)=-2(x-3)^3 ....or you can use the power rule for all functions cx^2=2cx

Please don't write nonsense: cx^2 is NOT equal to 2 cx. Yes, I know what you MEANT but that was what you wrote. Also the power rule does not apply to "all functions".

- #8

hawaiidude

- 41

- 0

ok mr smart ass

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