Directions: Consider the following function

f(x)

at

a=1

.\

f(x)=x-5

\ Part 1 - Difference Quotient\ If

h!=0

, then the difference quotient can be simplified into the form

Ah+B

. That is:\

(f(1+h)-f(1))/(h)=Ah+B

\ where

A

and

B

are constants. Find these constants:\

A=, and B=

\ (Note: It's possible for one or more of these constants to be 0.)\ Part 2 - Derivative\ Use the simplified expression from Part 1 to then calculate the derivative

f^(')(1)

:\

f^(')(1)=\\\\lim_(h->0)(f(1+h)-f(1))/(h)

\

f^(')(1)=

Directions: Consider the following function f(x) at a=1. f(x)=x5 Part 1 - Difference Quotient If h=0, then the difference quotient can be simplified into the form Ah+B. That is: hf(1+h)f(1)=Ah+B where A and B are constants. Find these constants: A=andB= (Note: It's possible for one or more of these constants to be 0.) Part 2 - Derivative Use the simplified expression from Part 1 to then calculate the derivative f(1) : f(1)=limh0hf(1+h)f(1) f(1)=