Given the function [tex]f(x)=-6+8 x^2[/tex], calculate the following values:
[tex]
\begin{array}{l}
f(a)=-6+8 a^2 \\
f(a+h)=\square \\
\frac{f(a+h)-f(a)}{h}=\square
\end{array}
[/tex]
Answer (1)
First, find f ( a + h ) by substituting a + h into the function: f ( a + h ) = − 6 + 8 ( a + h ) 2 = − 6 + 8 a 2 + 16 ah + 8 h 2 .
Next, calculate f ( a + h ) − f ( a ) = ( − 6 + 8 a 2 + 16 ah + 8 h 2 ) − ( − 6 + 8 a 2 ) = 16 ah + 8 h 2 .
Then, divide the result by h : h f ( a + h ) − f ( a ) = h 16 ah + 8 h 2 = 16 a + 8 h .
The final answers are f ( a + h ) = − 6 + 8 a 2 + 16 ah + 8 h 2 and h f ( a + h ) − f ( a ) = 16 a + 8 h .
Explanation
Understanding the Problem We are given the function f ( x ) = − 6 + 8 x 2 . We need to find f ( a + h ) and h f ( a + h ) − f ( a ) . We already know that f ( a ) = − 6 + 8 a 2 .
Calculating f(a+h) First, let's find f ( a + h ) . We substitute x = a + h into the function f ( x ) : f ( a + h ) = − 6 + 8 ( a + h ) 2
Expanding the Expression Now, let's expand the expression: f ( a + h ) = − 6 + 8 ( a 2 + 2 ah + h 2 ) = − 6 + 8 a 2 + 16 ah + 8 h 2
Calculating f(a+h) - f(a) Next, we need to find h f ( a + h ) − f ( a ) . First, let's find f ( a + h ) − f ( a ) :
f ( a + h ) − f ( a ) = ( − 6 + 8 a 2 + 16 ah + 8 h 2 ) − ( − 6 + 8 a 2 ) = 16 ah + 8 h 2
Calculating the Final Expression Now, we can find h f ( a + h ) − f ( a ) :
h f ( a + h ) − f ( a ) = h 16 ah + 8 h 2 = 16 a + 8 h
Final Answer Therefore, f ( a + h ) = − 6 + 8 a 2 + 16 ah + 8 h 2 and h f ( a + h ) − f ( a ) = 16 a + 8 h .
Examples
Understanding function transformations and difference quotients is crucial in calculus. For instance, consider the velocity of a car. If f ( t ) represents the position of the car at time t , then f ( a + h ) represents the position at a slightly later time a + h . The expression h f ( a + h ) − f ( a ) gives the average velocity of the car over the time interval [ a , a + h ] . As h approaches 0, this expression approaches the instantaneous velocity of the car at time a , which is the derivative f ′ ( a ) .
