Find local minimum of Y=X³−3X

Question

ScanSolve AI Tutor · Accepted Answer

Let's find that local minimum, Temitayo.

To find the local minimum of $Y = X^3 - 3X$ , we will use the first and second derivative tests.

Find the first derivative, $Y'$ .
$Y = X^3 - 3X$ $Y' = \frac{d}{dX}(X^3 - 3X) = 3X^2 - 3$
Set the first derivative to zero to find the critical points.
$3X^2 - 3 = 0$ $3X^2 = 3$ $X^2 = 1$ $X = \pm 1$
The critical points are $X = 1$ and $X = -1$ .
Find the second derivative, $Y''$ .
$Y'' = \frac{d}{dX}(3X^2 - 3) = 6X$
Use the second derivative test to classify the critical points.
- For $X = 1$ :
$Y''(1) = 6(1) = 6$
Since $Y''(1) > 0$ , there is a local minimum at $X = 1$ .
- For $X = -1$ :
$Y''(-1) = 6(-1) = -6$
Since $Y''(-1) < 0$ , there is a local maximum at $X = -1$ .
Find the $Y$ -coordinate for the local minimum at $X=1$ .
Substitute $X=1$ into the original function $Y = X^3 - 3X$ :
$Y = (1)^3 - 3(1)$ $Y = 1 - 3$ $Y = -2$
The local minimum occurs at the point $(1, -2)$ .

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