Question text [math: F(x)=x3−2x2+x−2(x−2)]\displaystyle F(x)=\frac{x^3-{2}x^2+x-{2}}{(x-{2})} is an antiderivative of [math: f(x)=2x]. Select the correct option. [select: (Clear my choice), It is true since the derivative of F(x) is equal to f(x)., It is false since the derivative of F(x) is not equal to f(x)] Check Question 1

Question

BlackTom AI · Accepted Answer

First, restating the setup helps ensure we’re analyzing the right expressions: F(x) is given as (x^3 − 2x^2 + x − 2)/(x − 2), and f(x) is 2x. The question asks whether F is an antiderivative of f, i.e., whether F'(x) = f(x) on an appropriate interval.

Option 1: 'It is true since the derivative of F(x) is equal to f(x).'
- To verify, simplify F(x) by performing polynomial division: divide x^3 − 2x^2 + x − 2 by x − 2. Using synthetic division with root 2 on the coefficients 1, −2, 1, −2 gives a quotient x^2 + 1 and a remainder 0, so F(x) = x^2 + 1 for all x ≠ 2.
- Differentiate: F'(x) = (x^2 + 1)' = 2x.
- Compare with f(x): f(x) = 2x. Therefore, for all x ≠ 2, F'(x) = f(x). Since F is not defined at x = 2 (the original expression has a denominator zero there), F can be considered an antiderivative of f on any interval that does not include x = 2. This supports the statement that F is an antiderivative of f on such intervals.

Option 2: 'It is false since the derivative of F(x) is not equal to f(x).'
- This would be true only if F'(x) differed from 2x on the domain where F is defined. However, as shown above, F'(x) = 2x for x ≠ 2, matching f(x). The claim in this option hinges on miscomputing the derivative or forgetting the domain restriction at x = 2.
- A common source of error is forgetting to simplify the quotient before differentiating or assuming F is defined at x = 2, which would imply F'(2) must equal f(2). Since F is not defined at x = 2, discussing equality at that single point is outside the standard interval-based definition of an antiderivative.

In summary, the key points are: simplifying F reveals F(x) = x^2 + 1 on its domain, its derivative is 2x, which matches f(x) for all x ≠ 2, and therefore the statement that F is an antiderivative of f is true on intervals not containing x = 2. The second option contradicts this by asserting no such equality holds, which is incorrect given the domain considerations and the derivative calculation.

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