Finding the Derivative of a Quadratic Function

[Chris_w]

find the value of the derivative of f(x)=-2(x-3)^2 at the point (2,-2)


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

 

[Hurkyl]

Basically, they just gave you an extra, useless bit of information. (Though that extra bit of information will be essential in problems in later sections)

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

 

[Chris_w]

ahh, tricky tricky :) Thanks hurkyl

 

[FrostScYthe]

f(x)=-2(x-3)^2
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

 

[hawaiidude]

or you can use the "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

 

[FrostScYthe]

[;)]

That just takes too long

 

[HallsofIvy]

hawaiidude posted or you can use the "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".

 

[hawaiidude]

ok mr smart ass