We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
336
2
avatar

f'(-4) if f(x) = x^2-x

So I'm calculating 

f´(-4) = lim(h goes against 0) (f(-4+h)-f(-4))/h

f´(-4) = lim(h goes against 0) ((-4+h)^2+4+4^2-4)/h

f´(-4) = lim(h goes against 0)(4^2-4^2-8h+h^2)/h

f´(-4) = lim(h goes against 0)h(-8+h)/h

f´(-4) = lim(h goes against 0) -8+h

f´(-4) = -8

 

I know the answer should be -9 since the diverate of f(x) = x^2-x should be 2x - 1 (-4*2-1) but I can't make it work :(

 May 29, 2017
 #1
avatar+100569 
+1

By definition, we have

 

[ (x + h)^2  - (x + h)  -  ( x^2 - x )  ]  /  h   =

 

[ x^2 + 2xh + h^2 - x - h  -  x^2  + x ]  /  h  =

 

[ 2xh  + h^2 -  h]  / h   =

 

[ (h)( 2x  + h  -  1 )]  / h   =     divide by h on top/bottom

 

[ 2x  + h  -  1 ]        let  h →  0     and we have

 

[ 2x  -  1 ]  =   f ' (x)

 

 

So

 

f '(-4)   =    2(-4)  -  1    =  - 8  -  1    =    - 9

 

 

 

cool cool cool

 May 29, 2017
 #2
avatar
0

Riight, forgot that you add (x+h) into all x terms in the function, thank you!

 May 29, 2017

4 Online Users