f(x) = x^3 - kx is increasing on R. Which values can k take?

Increasing & Decreasing Functions · Question 69 of 74

Question 69

MediumOriginal question

$f(x) = x^3 - kx$ is increasing on $\mathbb{R}$ . Which values can $k$ take?

k \le 0

Correct answerYour answer

k \ge 0

Correct answerYour answer

k = 0

onlyCorrect answerYour answerany real

k

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$f'(x) = 3x^2 - k$ must be $\ge 0$ for all $x$ ; its least value is $-k$ , at $x = 0$ .
So $-k \ge 0$ , that is $k \le 0$ .

Worth noting

$k = 0$ gives $x^3$ , which is increasing, but every negative $k$ works as well.

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

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