On $[1, 4]$ , where does $f(x) = x^2 - 6x + 10$ take its greatest value?

x = 3

, the critical pointCorrect answerYour answerAt

x = 1

, an end pointCorrect answerYour answerAt

x = 4

, an end pointCorrect answerYour answerIt has no greatest value on

[1, 4]

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

$f'(x) = 2x - 6 = 0$ at $x = 3$ , which is a minimum: $f(3) = 1$ .
The ends give $f(1) = 5$ and $f(4) = 2$ , so the greatest value, 5, is at $x = 1$ .

Worth noting

The only critical point is the least value here; the greatest value of a continuous function on a closed interval can sit at an end.

Ten questions from Maxima & Minima, at your own pace.

On [1,4][1, 4][1,4], where does f(x)=x2−6x+10f(x) = x^2 - 6x + 10f(x)=x2−6x+10 take its greatest value?