Find the difference quotient of [tex]$f$[/tex], that is, find [tex]$\frac{f(x+h)-f(x)}{h}, h \neq 0$[/tex], for the following function. Be sure to simplify.
[tex]$f(x)=x^2-5 x+1$[/tex]
[tex]$\frac{f(x+h)-f(x)}{h}=\square$[/tex]
Answer (1)
Find f ( x + h ) by substituting x + h into f ( x ) : f ( x + h ) = ( x + h ) 2 − 5 ( x + h ) + 1 = x 2 + 2 x h + h 2 − 5 x − 5 h + 1 .
Compute f ( x + h ) − f ( x ) : f ( x + h ) − f ( x ) = ( x 2 + 2 x h + h 2 − 5 x − 5 h + 1 ) − ( x 2 − 5 x + 1 ) = 2 x h + h 2 − 5 h .
Divide by h : h f ( x + h ) − f ( x ) = h 2 x h + h 2 − 5 h .
Simplify: h 2 x h + h 2 − 5 h = 2 x + h − 5 . The difference quotient is 2 x + h − 5 .
Explanation
Understanding the Problem We are given the function f ( x ) = x 2 − 5 x + 1 , and we need to find the difference quotient h f ( x + h ) − f ( x ) , where h = 0 .
Finding f(x+h) First, we need to find f ( x + h ) . We substitute x + h into the function f ( x ) : f ( x + h ) = ( x + h ) 2 − 5 ( x + h ) + 1 f ( x + h ) = x 2 + 2 x h + h 2 − 5 x − 5 h + 1
Computing f(x+h) - f(x) Next, we compute f ( x + h ) − f ( x ) : f ( x + h ) − f ( x ) = ( x 2 + 2 x h + h 2 − 5 x − 5 h + 1 ) − ( x 2 − 5 x + 1 ) f ( x + h ) − f ( x ) = x 2 + 2 x h + h 2 − 5 x − 5 h + 1 − x 2 + 5 x − 1 f ( x + h ) − f ( x ) = 2 x h + h 2 − 5 h
Dividing by h Now, we divide the result by h : h f ( x + h ) − f ( x ) = h 2 x h + h 2 − 5 h h f ( x + h ) − f ( x ) = h h ( 2 x + h − 5 )
Simplifying the Expression Finally, we simplify the expression by canceling h (since h = 0 ): h f ( x + h ) − f ( x ) = 2 x + h − 5
Final Answer Therefore, the difference quotient of f ( x ) = x 2 − 5 x + 1 is 2 x + h − 5 .
Examples
The difference quotient is a fundamental concept in calculus. It represents the average rate of change of a function over an interval of length h . In real-world applications, it can be used to approximate the velocity of an object at a specific time, given its position function. For example, if f ( x ) represents the position of a car at time x , then the difference quotient h f ( x + h ) − f ( x ) approximates the car's average velocity over the time interval [ x , x + h ] . As h approaches 0, the difference quotient approaches the instantaneous velocity at time x .
