For the function [tex]f(x)=x^3[/tex]

1) [tex]f(x+\Delta x)=f\left(x^3+\Delta x^3\right)=[/tex]
2) [tex]\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex]

First, find f ( x + Δ x ) by substituting ( x + Δ x ) into the function f ( x ) = x 3 , resulting in f ( x + Δ x ) = ( x + Δ x ) 3 = x 3 + 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 .
Next, compute the difference quotient Δ x f ( x + Δ x ) − f ( x ) ​ using the expressions for f ( x + Δ x ) and f ( x ) .
Simplify the difference quotient by canceling out terms and factoring, leading to Δ x f ( x + Δ x ) − f ( x ) ​ = 3 x 2 + 3 x Δ x + ( Δ x ) 2 .
The final expressions are f ( x + Δ x ) = x 3 + 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 and Δ x f ( x + Δ x ) − f ( x ) ​ = 3 x 2 + 3 x Δ x + ( Δ x ) 2 ​ .

Explanation

Problem Setup We are given the function f ( x ) = x 3 and asked to find expressions for f ( x + Δ x ) and Δ x f ( x + Δ x ) − f ( x ) ​ .

Finding f(x + Δx) First, we need to find f ( x + Δ x ) . Since f ( x ) = x 3 , we replace x with ( x + Δ x ) to get f ( x + Δ x ) = ( x + Δ x ) 3 .

Expanding (x + Δx)^3 Expanding ( x + Δ x ) 3 , we use the binomial theorem or simply multiply it out:


( x + Δ x ) 3 = ( x + Δ x ) ( x + Δ x ) ( x + Δ x ) = ( x 2 + 2 x Δ x + ( Δ x ) 2 ) ( x + Δ x ) = x 3 + 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 .
So, f ( x + Δ x ) = x 3 + 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 .

Finding the Difference Quotient Next, we need to find Δ x f ( x + Δ x ) − f ( x ) ​ . We know that f ( x + Δ x ) = x 3 + 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 and f ( x ) = x 3 . Substituting these into the expression, we get:

Δ x f ( x + Δ x ) − f ( x ) ​ = Δ x ( x 3 + 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 ) − x 3 ​ .

Simplifying the Expression Simplifying the expression, we have:

Δ x 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 ​ .
We can factor out Δ x from the numerator:
Δ x Δ x ( 3 x 2 + 3 x Δ x + ( Δ x ) 2 ) ​ .
Canceling Δ x from the numerator and denominator (assuming Δ x  = 0 ), we get:
3 x 2 + 3 x Δ x + ( Δ x ) 2 .

Final Answer Therefore, f ( x + Δ x ) = x 3 + 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 and Δ x f ( x + Δ x ) − f ( x ) ​ = 3 x 2 + 3 x Δ x + ( Δ x ) 2 .

Examples
Understanding how a function changes as its input changes is fundamental in many fields. For instance, in physics, if f ( x ) represents the position of an object at time x , then Δ x f ( x + Δ x ) − f ( x ) ​ represents the average velocity of the object over the time interval Δ x . As Δ x approaches 0, this expression gives the instantaneous velocity at time x , which is a crucial concept in mechanics. Similarly, in economics, this concept can be used to analyze marginal cost or marginal revenue, helping businesses make informed decisions about production levels.

Answered by GinnyAnswer | 2025-07-05

We found that f ( x + Δ x ) = x 3 + 3 x 2 Δ x + 3 x ( Δ x ) 2 + ( Δ x ) 3 and the difference quotient is 3 x 2 + 3 x Δ x + ( Δ x ) 2 .
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Answered by Anonymous | 2025-08-07