Find each of the following for [tex]f(x)=4 x^2-6 x+5[/tex].
(a) [tex]f(x+h)[/tex]
(b) [tex]f(x+h)-f(x)[/tex]
(c) [tex]\frac{f(x+h)-f(x)}{h}[/tex]
Answer (1)
Find f ( x + h ) by substituting x + h into f ( x ) : f ( x + h ) = 4 ( x + h ) 2 − 6 ( x + h ) + 5 = 4 x 2 + 8 x h + 4 h 2 − 6 x − 6 h + 5 .
Find f ( x + h ) − f ( x ) : f ( x + h ) − f ( x ) = ( 4 x 2 + 8 x h + 4 h 2 − 6 x − 6 h + 5 ) − ( 4 x 2 − 6 x + 5 ) = 8 x h + 4 h 2 − 6 h .
Find h f ( x + h ) − f ( x ) : h f ( x + h ) − f ( x ) = h 8 x h + 4 h 2 − 6 h = 8 x + 4 h − 6 .
The final answers are: f ( x + h ) = 4 x 2 + 8 x h + 4 h 2 − 6 x − 6 h + 5 , f ( x + h ) − f ( x ) = 8 x h + 4 h 2 − 6 h , and h f ( x + h ) − f ( x ) = 8 x + 4 h − 6 .
Explanation
Understanding the Problem We are given the function f ( x ) = 4 x 2 − 6 x + 5 and we need to find expressions for f ( x + h ) , f ( x + h ) − f ( x ) , and h f ( x + h ) − f ( x ) .
Finding f(x+h) First, we find f ( x + h ) by substituting x + h for x in the expression for f ( x ) :
f ( x + h ) = 4 ( x + h ) 2 − 6 ( x + h ) + 5
Expanding this, we get:
f ( x + h ) = 4 ( x 2 + 2 x h + h 2 ) − 6 ( x + h ) + 5 f ( x + h ) = 4 x 2 + 8 x h + 4 h 2 − 6 x − 6 h + 5
Finding f(x+h) - f(x) Next, we find f ( x + h ) − f ( x ) by subtracting f ( x ) from f ( x + h ) :
f ( x + h ) − f ( x ) = ( 4 x 2 + 8 x h + 4 h 2 − 6 x − 6 h + 5 ) − ( 4 x 2 − 6 x + 5 )
Simplifying, we get:
f ( x + h ) − f ( x ) = 4 x 2 + 8 x h + 4 h 2 − 6 x − 6 h + 5 − 4 x 2 + 6 x − 5 f ( x + h ) − f ( x ) = 8 x h + 4 h 2 − 6 h
Finding [f(x+h) - f(x)] / h Finally, we find h f ( x + h ) − f ( x ) by dividing the result from the previous step by h :
h f ( x + h ) − f ( x ) = h 8 x h + 4 h 2 − 6 h
Factoring out h from the numerator, we get:
h f ( x + h ) − f ( x ) = h h ( 8 x + 4 h − 6 )
Canceling h , we get:
h f ( x + h ) − f ( x ) = 8 x + 4 h − 6
Final Answer Therefore, we have:
(a) f ( x + h ) = 4 x 2 + 8 x h + 4 h 2 − 6 x − 6 h + 5 (b) f ( x + h ) − f ( x ) = 8 x h + 4 h 2 − 6 h (c) h f ( x + h ) − f ( x ) = 8 x + 4 h − 6
Examples
Understanding how a function changes as its input changes is a fundamental concept in calculus. For example, if f ( x ) represents the position of a car at time x , then f ( x + h ) − f ( x ) represents the change in position over a time interval of length h . The expression h f ( x + h ) − f ( x ) represents the average velocity of the car over that time interval. As h approaches 0, this expression approaches the instantaneous velocity of the car at time x .
