Assertion (A): On $[-2, 2]$ , the greatest value of $f(x) = x^3 - 3x$ is 2.
Reason (R): On a closed interval, a local maximum value is always the greatest value.

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 C

Correct answer: C

$f'(x) = 3(x - 1)(x + 1)$ ; comparing $f(-2) = -2$ , $f(-1) = 2$ , $f(1) = -2$ and $f(2) = 2$ , the greatest value is 2: A is true.
R is false: on $[-2, 3]$ the same local maximum value 2 is beaten by the end value $f(3) = 18$ .

Concept: Local maximum or greatest value
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsGreatest and least values on an interval

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

Assertion (A): On [−2,2][-2, 2][−2,2], the greatest value of f(x)=x3−3xf(x) = x^3 - 3xf(x)=x3−3x is 2.Reason (R): On a closed interval, a local maximum value is always the greatest value.

Explanation

Assertion (A): On $[-2, 2]$ , the greatest value of $f(x) = x^3 - 3x$ is 2.
Reason (R): On a closed interval, a local maximum value is always the greatest value.