Assertion (A): f(x) = x^3 + 3x^2 + 3x + 1 is increasing on R. Reason (R): f'(x) = 3(x + 1)^2 ≥ 0 for every x.

Question 68

MediumOriginal question

Assertion (A): $f(x) = x^3 + 3x^2 + 3x + 1$ is increasing on $\mathbb{R}$ .
Reason (R): $f'(x) = 3(x + 1)^2 \ge 0$ for every $x$ .

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

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$f'(x) = 3x^2 + 6x + 3 = 3(x + 1)^2$ , never negative: R is true.
A derivative that is never negative, and 0 only at $x = -1$ , makes $f$ increasing on $\mathbb{R}$ (indeed $f = (x + 1)^3$ ): A is true, and R explains it.

Concept: A perfect-cube derivative test
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsConditions on a constant

Practise this chapter →

Ten questions from Increasing & Decreasing Functions, at your own pace.

Assertion (A): f(x)=x3+3x2+3x+1f(x) = x^3 + 3x^2 + 3x + 1f(x)=x3+3x2+3x+1 is increasing on R\mathbb{R}R.Reason (R): f′(x)=3(x+1)2≥0f'(x) = 3(x + 1)^2 \ge 0f′(x)=3(x+1)2≥0 for every xxx.

Explanation

Assertion (A): $f(x) = x^3 + 3x^2 + 3x + 1$ is increasing on $\mathbb{R}$ .
Reason (R): $f'(x) = 3(x + 1)^2 \ge 0$ for every $x$ .