Assertion (A): f(x) = e^-x^2 is strictly increasing on (-∞, 0). Reason (R): f'(x) = -2xe^-x^2, which is positive for every x < 0.

Increasing & Decreasing Functions · Question 25 of 74

Question 25

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Assertion (A): $f(x) = e^{-x^2}$ is strictly increasing on $(-\infty, 0)$ .
Reason (R): $f'(x) = -2xe^{-x^2}$ , which is positive for every $x < 0$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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CorrectNot quiteAnswer A

Correct answer: A

R: for $x < 0$ , $-2x > 0$ and $e^{-x^2} > 0$ , so $f'(x) > 0$ . R is true.
A positive derivative on the interval is exactly what makes $f$ strictly increasing there, so R explains A.

Concept: The derivative test
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Assertion (A): f(x)=e−x2f(x) = e^{-x^2}f(x)=e−x2 is strictly increasing on (−∞,0)(-\infty, 0)(−∞,0).Reason (R): f′(x)=−2xe−x2f'(x) = -2xe^{-x^2}f′(x)=−2xe−x2, which is positive for every x<0x < 0x<0.

Explanation

Assertion (A): $f(x) = e^{-x^2}$ is strictly increasing on $(-\infty, 0)$ .
Reason (R): $f'(x) = -2xe^{-x^2}$ , which is positive for every $x < 0$ .