If the function [tex]f(x)=x^2+7[/tex] then evaluate:

1) [tex]f(3 a)=[/tex]
2) [tex]f(b-1)=[/tex]
3) [tex]\frac{f(x+\Delta x)-f(x)}{\Delta x} \quad[/tex] where [tex]\Delta x \neq 0[/tex]. Choose the correct answer: A) [tex]2 x+\Delta x[/tex] B) [tex]x+\Delta x[/tex] C) [tex]2 x-\Delta x[/tex]

Evaluate f ( 3 a ) by substituting 3 a into the function: f ( 3 a ) = ( 3 a ) 2 + 7 = 9 a 2 + 7 .
Evaluate f ( b − 1 ) by substituting b − 1 into the function: f ( b − 1 ) = ( b − 1 ) 2 + 7 = b 2 − 2 b + 8 .
Evaluate Δ x f ( x + Δ x ) − f ( x ) ​ by finding f ( x + Δ x ) = ( x + Δ x ) 2 + 7 and simplifying the expression to 2 x + Δ x .
The final answers are f ( 3 a ) = 9 a 2 + 7 , f ( b − 1 ) = b 2 − 2 b + 8 , and Δ x f ( x + Δ x ) − f ( x ) ​ = 2 x + Δ x ​ .

Explanation

Problem Analysis We are given the function f ( x ) = x 2 + 7 and asked to evaluate three expressions: f ( 3 a ) , f ( b − 1 ) , and Δ x f ( x + Δ x ) − f ( x ) ​ . We will evaluate each expression by substituting the given input into the function and simplifying.

Evaluating f(3a) First, we evaluate f ( 3 a ) . We substitute 3 a for x in the function: f ( 3 a ) = ( 3 a ) 2 + 7 = 9 a 2 + 7

Evaluating f(b-1) Next, we evaluate f ( b − 1 ) . We substitute b − 1 for x in the function: f ( b − 1 ) = ( b − 1 ) 2 + 7 Expanding the square, we get: f ( b − 1 ) = b 2 − 2 b + 1 + 7 = b 2 − 2 b + 8

Evaluating the Difference Quotient Finally, we evaluate Δ x f ( x + Δ x ) − f ( x ) ​ . First, we find f ( x + Δ x ) : f ( x + Δ x ) = ( x + Δ x ) 2 + 7 = x 2 + 2 x Δ x + ( Δ x ) 2 + 7 Then, we compute the difference quotient: Δ x f ( x + Δ x ) − f ( x ) ​ = Δ x ( x 2 + 2 x Δ x + ( Δ x ) 2 + 7 ) − ( x 2 + 7 ) ​ = Δ x 2 x Δ x + ( Δ x ) 2 ​ Factoring out Δ x from the numerator, we get: Δ x Δ x ( 2 x + Δ x ) ​ = 2 x + Δ x

Identifying the Correct Option Comparing the result 2 x + Δ x with the given options, we see that it matches option A.

Final Answer Therefore, the answers are:



f ( 3 a ) = 9 a 2 + 7
f ( b − 1 ) = b 2 − 2 b + 8
Δ x f ( x + Δ x ) − f ( x ) ​ = 2 x + Δ x , which corresponds to option A.

Examples
Understanding function evaluation is crucial in many real-world applications. For instance, in physics, if f ( t ) represents the distance an object falls in time t , then f ( 5 ) would tell us how far the object has fallen after 5 seconds. Similarly, in economics, if C ( x ) represents the cost of producing x items, then C ( 100 ) would be the cost of producing 100 items. Evaluating expressions like these helps us make predictions and understand the behavior of systems.

Answered by GinnyAnswer | 2025-07-05

The evaluations of the function f ( x ) = x 2 + 7 yield f ( 3 a ) = 9 a 2 + 7 , f ( b − 1 ) = b 2 − 2 b + 8 , and the difference quotient simplifies to 2 x + Δ x , which matches option A.
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Answered by Anonymous | 2025-07-07