Find local minimum of Y=X³-3X

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.

Step 1: Find the first derivative, $Y'$. $Y = X^3 - 3X$ $Y' = \frac{d}{dX}(X^3 - 3X) = 3X^2 - 3$

Step 2: 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 $X = -1$.

Step 3: Find the second derivative, $Y''$. $Y'' = \frac{d}{dX}(3X^2 - 3) = 6X$

Step 4: 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 $X = -1$.

Step 5: 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 $\boxed{(1, -2)}$.

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