Let f(x) = -2 x + 5 x^2. Then the expression (f(x+h)-f(x))/h?

(1 pt) Let f(x) = -2 x + 5 x^2. Then the expression
(f(x+h)-f(x))/h
can be written in the form Ah + Bx + C, where A, B, and C are constants. (Note: It's possible for one or more of these constants to be 0.) Find the constants.
A =
B =
C =

Use your answer from above to find lim has h->0 (f(x+h)-f(x))/h =_

Finally, find each of the following:

f'(1) =
f'(2) =
f'(3) =

2 Answers
Shwetank Mauria
Jan 19, 2018

#A=5#, #B=10# and #C=-2#
#f'(1)=8#, #f'(2)=18# and #f'(3)=28#

Explanation:

As # f(x)=-2x+5x^2#

#f(x+h)=-2(x+h)+5(x+h)^2#

= #-2x-2h+5(x^2+2hx+h^2)#

= #-2x-2h+5x^2+10hx+5h^2#

and #(f(x+h)-f(x))/h=(-2x-2h+5x^2+10hx+5h^2-(-2x+5x^2))/h#

= #(-2x-2h+5x^2+10hx+5h^2+2x-5x^2)/h#

= #(10hx+5h^2-2h)/h#

= #10x+5h-2# or #5h+10x-2#

Comparing it with #Ah+Bx+C#, we have

#A=5#, #B=10# and #C=-2#

and #f'(x)=lim_(h->0)(f(x+h)-f(x))/h=10x-2#

and #f'(1)=10xx1-2=8#, #f'(2)=10xx2-2=18# and #f'(3)=10xx3-2=28#

Ratnaker Mehta
Jan 19, 2018

#"Part I : "A=5, B=10, C=-2#.

#"Part II : "10x-2#.

#"Part III : "8, 18, 28#.

Explanation:

Part I :

#f(x)=-2x+5x^2#,

#rArr f(x+h)=-2(x+h)+5(x+h)^2#,

#=-2x-2h+5x^2+10xh+5h^2#.

#:. (f(x+h)-f(x))/h#,

#=1/h{(-2x-2h+5x^2+10xh+5h^2)-(-2x+5x^2)}#,

#=1/h{-2h+10xh+5h^2}#,

#rArr (f(x+h)-f(x))/h=5h+10x-2#.

#:. (f(x+h)-f(x))/h=Ah+Bx+C#,

#rArr A=5, B=10, C=-2#.

Part II :

#lim_(h to 0)(f(x+h)-f(x))/h#,

#=lim_(h to 0)5h+10x-2#,

#=5(0)+10x-2#.

#:. lim_(h to 0)(f(x+h)-f(x))/h=10x-2#.

Part III :

Recall that, #f'(x)=lim_(h to 0)(f(x+h)-f(x))/h#.

#:. f'(x)=10x-2#.

#:. f'(1)=10(1)-2=8#,

#f'(2)=18, and, f'(3)=28#.

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