Finding Maximum and Minimum Points of a Function

To determine where a function has local maximum or minimum points, we use calculus. The key steps involve finding the first and second derivatives of the function.

The given function is:

\( Y = X^3 - 12X^2 + 36X + 8 \)

Calculating the First Derivative

The first derivative of the function, denoted \( Y' \) or \( \frac{dY}{dX} \), tells us about the slope of the tangent line to the curve at any point \( X \). Local maximum or minimum points occur where the slope is zero.

\( Y' = \frac{d}{dX}(X^3 - 12X^2 + 36X + 8) \)

Using the power rule (\( \frac{d}{dX}(X^n) = nX^{n-1} \)) and the rule for constants (\( \frac{d}{dX}(c) = 0 \)):

\( Y' = 3X^{3-1} - 12 \cdot 2X^{2-1} + 36 \cdot 1X^{1-1} + 0 \)

\( Y' = 3X^2 - 24X + 36 \)

Finding Critical Points

Critical points are potential locations for local maximum or minimum values. They occur where the first derivative is equal to zero or undefined. For polynomial functions, the first derivative is always defined. So, we set \( Y' = 0 \):

\( 3X^2 - 24X + 36 = 0 \)

We can divide the entire equation by 3 to simplify:

\( X^2 - 8X + 12 = 0 \)

This is a quadratic equation. We can solve it by factoring:

\( (X - 2)(X - 6) = 0 \)

Setting each factor to zero gives the critical points:

• \( X - 2 = 0 \implies X = 2 \)
• \( X - 6 = 0 \implies X = 6 \)

The critical points are \( X = 2 \) and \( X = 6 \).

Calculating the Second Derivative

The second derivative, denoted \( Y'' \) or \( \frac{d^2Y}{dX^2} \), tells us about the concavity of the function. We use it to classify the critical points found in the previous step.

\( Y'' = \frac{d}{dX}(3X^2 - 24X + 36) \)

Using the power rule again:

\( Y'' = 3 \cdot 2X^{2-1} - 24 \cdot 1X^{1-1} + 0 \)

\( Y'' = 6X - 24 \)

Applying the Second Derivative Test

We evaluate the second derivative at each critical point:

At \( X = 2 \):

\( Y''(2) = 6(2) - 24 \)

\( Y''(2) = 12 - 24 \)

\( Y''(2) = -12 \)

Since \( Y''(2) < 0 \), the function has a local maximum at \( X = 2 \).

At \( X = 6 \):

\( Y''(6) = 6(6) - 24 \)

\( Y''(6) = 36 - 24 \)

\( Y''(6) = 12 \)

Since \( Y''(6) > 0 \), the function has a local minimum at \( X = 6 \).

Based on this analysis, the function \( Y \) is minimum at \( X = 6 \) and maximum at \( X = 2 \).

Evaluating the Options

Let's check the given options:

• Option 1: Y is maximum at X = 6, and minimum at X = 2. This contradicts our finding.
• Option 2: Y is minimum at X = 6, and maximum at X = 2. This matches our finding.
• Option 3: Y has two inflexion points at X = 2 and X = 6. Inflection points occur where the concavity changes, typically where the second derivative is zero or undefined. Our analysis shows X=2 and X=6 are extrema, not inflection points.
• Option 4: Point of optimality is indeterminate. We were able to clearly determine the local maximum and minimum points.

Therefore, the correct statement is that Y is minimum at \( X = 6 \) and maximum at \( X = 2 \).

Revision Table: Finding Extrema

Additional Information: Inflection Points and Concavity

While this question focuses on maximum and minimum points, it's useful to understand inflection points, which are mentioned in one of the options.

• Inflection Point: A point on the curve where the concavity changes.
• Concavity: Describes whether the curve is bending upwards (concave up) or downwards (concave down).

Concavity is determined by the sign of the second derivative \( Y'' \):

• If \( Y'' > 0 \) on an interval, the function is concave up on that interval.
• If \( Y'' < 0 \) on an interval, the function is concave down on that interval.

Inflection points occur where \( Y'' = 0 \) or is undefined, and where the sign of \( Y'' \) changes around that point.

For our function \( Y'' = 6X - 24 \). Setting \( Y'' = 0 \):

\( 6X - 24 = 0 \)

\( 6X = 24 \)

\( X = 4 \)

Now, check the sign of \( Y'' \) around \( X = 4 \):

• For \( X < 4 \) (e.g., \( X = 3 \)): \( Y''(3) = 6(3) - 24 = 18 - 24 = -6 < 0 \). The function is concave down.
• For \( X > 4 \) (e.g., \( X = 5 \)): \( Y''(5) = 6(5) - 24 = 30 - 24 = 6 > 0 \). The function is concave up.

Since the concavity changes at \( X = 4 \), there is an inflection point at \( X = 4 \). This confirms that options stating inflection points at X=2 and X=6 are incorrect.