Find f ( x + h ) by substituting x + h into f ( x ) , resulting in f ( x + h ) = ( x + h ) 2 + 7 .
Substitute f ( x + h ) and f ( x ) into the difference quotient: h f ( x + h ) − f ( x ) = h (( x + h ) 2 + 7 ) − ( x 2 + 7 ) .
Simplify the expression by expanding, canceling terms, and factoring: h 2 x h + h 2 = h h ( 2 x + h ) .
Cancel the common factor h , yielding the simplified difference quotient: 2 x + h .
Explanation
Understanding the Problem We are given the function f ( x ) = x 2 + 7 and asked to simplify the difference quotient h f ( x + h ) − f ( x ) , where h = 0 . This quotient is a fundamental concept in calculus, representing the average rate of change of the function f ( x ) over an interval of length h .
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 + 7
Expanding (x+h)^2 Next, we expand ( x + h ) 2 :
( x + h ) 2 = x 2 + 2 x h + h 2
Substituting the expansion So, we have: f ( x + h ) = x 2 + 2 x h + h 2 + 7
Substituting into the difference quotient Now, we substitute f ( x + h ) and f ( x ) into the difference quotient: h f ( x + h ) − f ( x ) = h ( x 2 + 2 x h + h 2 + 7 ) − ( x 2 + 7 )
Simplifying the numerator Simplify the numerator by canceling terms: h x 2 + 2 x h + h 2 + 7 − x 2 − 7 = h 2 x h + h 2
Factoring out h Factor out h from the numerator: h h ( 2 x + h )
Canceling h Cancel the common factor h from the numerator and denominator, since h = 0 :
2 x + h
Final Answer Therefore, the simplified difference quotient is 2 x + h .
Examples
The difference quotient is used to find the derivative of a function, which represents the instantaneous rate of change. For example, if f ( x ) represents the position of a car at time x , then the derivative f ′ ( x ) represents the car's velocity at time x . The difference quotient approximates this velocity over a small time interval h . Understanding difference quotients is crucial in physics, engineering, and economics for analyzing rates of change and making predictions.