Aspire Faculty ID #13529 · Topic: JAMIA MILLIA ISLAMIA MCA 2024 · Just now
JAMIA MILLIA ISLAMIA MCA 2024

The absolute maximum value of $y = x^3 - 3x + 2$ in $0 \le x \le 2$ is

Solution

$y' = 3x^2 - 3 = 0 \Rightarrow x = 1$ Now, $y(0) = 2$, $y(1) = 0$, $y(2) = 8 - 6 + 2 = 4$ Hence, maximum value = 4.
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