If [tex]f(x)=x^3[/tex] then find [tex]\frac{f(x)-f(2)}{x-2}[/tex]
A. [tex]x^2+2 x+4[/tex] B. [tex]x^2+4 x+4[/tex] C. [tex]x^2+2 x[/tex] D. [tex]x^2-2 x[/tex]
Answer (1)
Calculate f ( 2 ) : f ( 2 ) = 2 3 = 8 .
Substitute into the expression: x − 2 f ( x ) − f ( 2 ) = x − 2 x 3 − 8 .
Factor the numerator using the difference of cubes: x 3 − 8 = ( x − 2 ) ( x 2 + 2 x + 4 ) .
Simplify the expression by canceling the common factor: x − 2 ( x − 2 ) ( x 2 + 2 x + 4 ) = x 2 + 2 x + 4 . The final answer is x 2 + 2 x + 4 .
Explanation
Understanding the Problem We are given the function f ( x ) = x 3 and asked to find the expression x − 2 f ( x ) − f ( 2 ) . This looks like we're heading towards finding a derivative, but let's just simplify the expression directly.
Calculate f(2) First, let's find f ( 2 ) . Since f ( x ) = x 3 , we have f ( 2 ) = 2 3 = 8 .
Substitute into the Expression Now, substitute f ( x ) = x 3 and f ( 2 ) = 8 into the expression: x − 2 f ( x ) − f ( 2 ) = x − 2 x 3 − 8 .
Factor the Numerator We can factor the numerator x 3 − 8 using the difference of cubes factorization, which is a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) . In this case, a = x and b = 2 , so we have x 3 − 8 = x 3 − 2 3 = ( x − 2 ) ( x 2 + 2 x + 4 ) .
Substitute Factored Form Now, substitute the factored form back into the expression: x − 2 ( x − 2 ) ( x 2 + 2 x + 4 ) .
Simplify the Expression We can cancel the common factor ( x − 2 ) from the numerator and the denominator, provided that x = 2 : x − 2 ( x − 2 ) ( x 2 + 2 x + 4 ) = x 2 + 2 x + 4.
Final Answer Therefore, the simplified expression is x 2 + 2 x + 4 . Comparing this to the given options, we see that it matches option A.
Examples
Understanding how to simplify expressions like this is useful in calculus when finding derivatives using the limit definition. For example, the derivative of f ( x ) = x 3 at x = 2 can be found by evaluating the limit as x approaches 2 of the expression x − 2 f ( x ) − f ( 2 ) , which we just simplified. This type of simplification is also useful in physics when calculating average rates of change.
