For what values of $k$ is $f(x) = x^2 - kx + 4$ strictly increasing on $(3, \infty)$ ?

k \ge 6

Correct answerYour answer

k \le 3

Correct answerYour answer

k \le 6

Correct answerYour answer

k \le -6

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$f'(x) = 2x - k$ , which is positive for $x > \frac{k}{2}$ .
The whole interval $(3, \infty)$ must lie in $x > \frac{k}{2}$ , so $\frac{k}{2} \le 3$ , that is $k \le 6$ .

Worth noting

$k \le 3$ forgets that the turning point is at $\frac{k}{2}$ , not at $k$ .

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

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

For what values of kkk is f(x)=x2−kx+4f(x) = x^2 - kx + 4f(x)=x2−kx+4 strictly increasing on (3,∞)(3, \infty)(3,∞)?