Suppose
𝑔
′
(
𝑥
)
=
𝑥
𝑥
2
+
1
. Where are the inflection points for
𝑔
(
𝑥
)
?

Question

Suppose 
𝑔
′
(
𝑥
)
=
𝑥
𝑥
2
+
1
. Where are the inflection points for 
𝑔
(
𝑥
)
?

BlackTom AI · Accepted Answer

Question restatement: We are asked to find the inflection points of g(x) given that g′(x) = x/(x^2 + 1). To locate inflection points, we examine where g″(x) changes sign or is undefined, since inflection points occur where the concavity of g changes.

Step 1: Compute g″(x) from g′(x).
- Use the quotient rule for f(x) = x and h(x) = x^2 + 1:
  g″(x) = [ (x^2 + 1)·1 − x·(2x) ] / (x^2 + 1)^2
  = (x^2 + 1 − 2x^2) / (x^2 + 1)^2
  = (1 − x^2) / (x^2 + 1)^2.

Step 2: Find where g″(x) = 0 or is undefined.
- The denominator (x^2 + 1)^2 is always positive for all real x, hence never undefined.
- The numerator 1 − x^2 equals zero when x^2 = 1, i.e., x = −1 or x = 1.

Step 3: Determine sign changes to confirm inflection points.
- Since the denominator is always positive, the sign of g″(x) is governed by 1 − x^2.
- As x passes through −1 and 1, the factor (1 − x^2) changes sign (from positive to negative or negative to positive), indicating a change in concavity at those x-values.

Conclusion (based on standard inflection-point criteria): the inflection points occur at x = −1 and x = 1. Therefore, the stated answer −1, 1 is correct.

Note on options: The provided data shows no list of multiple-choice options to analyze, only the stated correct pair (−1, 1). In practice, if options included '−1, 1' among others, this analysis would justify selecting '−1, 1' as the inflection points by demonstrating where g″(x) changes sign.

MATH-112-301-001 Unproctored Midcourse Exam 3 Practice Exam 1

Suppose 𝑔 ′ ( 𝑥 ) = 𝑥 𝑥 2 + 1 . Where are the inflection points for 𝑔 ( 𝑥 ) ?

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