Find each of the following for [tex]f(x)=5 x^2-8 x+6[/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]
(a) [tex]f(x+h)=[/tex] $\square$
Answer (1)
Find f ( x + h ) by substituting x + h into f ( x ) : f ( x + h ) = 5 ( x + h ) 2 − 8 ( x + h ) + 6 = 5 x 2 + 10 x h + 5 h 2 − 8 x − 8 h + 6 .
Find f ( x + h ) − f ( x ) : f ( x + h ) − f ( x ) = ( 5 x 2 + 10 x h + 5 h 2 − 8 x − 8 h + 6 ) − ( 5 x 2 − 8 x + 6 ) = 10 x h + 5 h 2 − 8 h .
Find h f ( x + h ) − f ( x ) : h f ( x + h ) − f ( x ) = h 10 x h + 5 h 2 − 8 h = 10 x + 5 h − 8 .
The expression for f ( x + h ) is 5 x 2 + 10 x h + 5 h 2 − 8 x − 8 h + 6 .
Explanation
Understanding the Problem We are given the function f ( x ) = 5 x 2 − 8 x + 6 and we need to find 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 ) = 5 ( x + h ) 2 − 8 ( x + h ) + 6
Expanding the terms Expanding the terms, we get: f ( x + h ) = 5 ( x 2 + 2 x h + h 2 ) − 8 ( x + h ) + 6
Distributing the constants Distributing the constants, we have: f ( x + h ) = 5 x 2 + 10 x h + 5 h 2 − 8 x − 8 h + 6
Result for f(x+h) So, f ( x + h ) = 5 x 2 + 10 x h + 5 h 2 − 8 x − 8 h + 6 .
Finding f(x+h) - f(x) Next, we find f ( x + h ) − f ( x ) :
f ( x + h ) − f ( x ) = ( 5 x 2 + 10 x h + 5 h 2 − 8 x − 8 h + 6 ) − ( 5 x 2 − 8 x + 6 ) f ( x + h ) − f ( x ) = 5 x 2 + 10 x h + 5 h 2 − 8 x − 8 h + 6 − 5 x 2 + 8 x − 6
Simplifying the expression Simplifying, we get: f ( x + h ) − f ( x ) = 10 x h + 5 h 2 − 8 h
Finding the final expression Finally, we find h f ( x + h ) − f ( x ) :
h f ( x + h ) − f ( x ) = h 10 x h + 5 h 2 − 8 h h f ( x + h ) − f ( x ) = h h ( 10 x + 5 h − 8 )
Simplifying the final expression Cancelling h , we get: h f ( x + h ) − f ( x ) = 10 x + 5 h − 8
Examples
Understanding function transformations is crucial in many fields. For example, in physics, understanding how a function changes with respect to time (similar to adding 'h' to 'x') can help model the velocity and acceleration of an object. In economics, it can help analyze how costs or profits change with respect to production levels. These concepts provide a foundation for more advanced mathematical modeling and analysis.
