Question

Find the relative maxima and minima of \(y\) by the second-derivative test: (a) \(y=-2 x^{2}+8 x+25\) (b) \(y=x^{3}+6 x^{2}+9\) (c) \(y=\frac{1}{3} x^{3}-3 x^{2}+5 x+3\) (d) \(y=\frac{2 x}{1-2 x} \quad\left(x \neq \frac{1}{2}\right)\)

Short answer

(a) Max: \(x = 2\); (b) Min: \(x = 0\), Max: \(x = -4\); (c) Max: \(x = 1\), Min: \(x = 5\); (d) Max: \(x = \frac{1}{4}\).

Step-by-step solution

Identify First Derivative for Function (a)

The function given is \(y=-2x^2+8x+25\). First, find the first derivative, \( y' \), using the power rule. The derivative of \(-2x^2\) is \(-4x\), and the derivative of \(8x\) is \(8\), giving us: \[ y' = -4x + 8 \]

Find Critical Points for Function (a)

Set the first derivative \( y' = -4x + 8 \) equal to zero and solve for \( x \): \[ -4x + 8 = 0 \] \[ -4x = -8 \] \[ x = 2 \] This is the critical point where potential maxima or minima can occur.

Determine Second Derivative for Concavity (a)

Find the second derivative \( y'' \) of \(y=-2x^2+8x+25\) from \( y' = -4x + 8 \). Differentiating \(-4x + 8\) gives: \[ y'' = -4 \]

Apply the Second-Derivative Test (a)

With \( y'' = -4 \), which is less than zero, the function is concave down at the critical point \( x = 2 \). Thus, \( x = 2 \) is a point of relative maximum.

Identify First Derivative for Function (b)

The function is \(y = x^3 + 6x^2 + 9\). First, find its first derivative: \[ y' = 3x^2 + 12x \] using the power rule on \(x^3\) and \(6x^2\).

Find Critical Points for Function (b)

Set \( y' = 3x^2 + 12x \) equal to zero to find critical points: \[ 3x^2 + 12x = 0 \] Factor out \(3x\): \[ 3x(x + 4) = 0 \] This gives two critical points: \( x = 0 \) and \( x = -4 \).

Determine Second Derivative for Concavity (b)

Find the second derivative \( y'' \): \[ y'' = 6x + 12 \] Calculate \( y'' \) at each critical point. For \( x = 0 \), \( y''(0) = 6(0) + 12 = 12 \). For \( x = -4 \), \( y''(-4) = 6(-4) + 12 = -12 \).

Apply the Second-Derivative Test (b)

For \( x = 0 \), \( y'' > 0 \) indicates a relative minimum. For \( x = -4 \), \( y'' < 0 \) indicates a relative maximum.

Identify First Derivative for Function (c)

The function is \(y = \frac{1}{3}x^3 - 3x^2 + 5x + 3\). First, find its first derivative: \[ y' = x^2 - 6x + 5 \]

Find Critical Points for Function (c)

Set \( y' = x^2 - 6x + 5 \) equal to zero: \[ x^2 - 6x + 5 = 0 \] Factor: \((x - 1)(x - 5) = 0\), giving \( x = 1 \) and \( x = 5 \).

Determine Second Derivative for Concavity (c)

Find the second derivative \( y'' \): \[ y'' = 2x - 6 \] Calculate \( y''(1) = 2(1) - 6 = -4 \); \( y''(5) = 2(5) - 6 = 4 \).

Apply the Second-Derivative Test (c)

At \( x = 1 \), \( y'' < 0 \) indicates a relative maximum. At \( x = 5 \), \( y'' > 0 \) indicates a relative minimum.

Identify First Derivative for Function (d)

The function is \(y = \frac{2x}{1-2x}\). First, apply the quotient rule to find the derivative: \[ y' = \frac{2(1-2x) - 4x}{(1-2x)^2} = \frac{2 - 8x}{(1-2x)^2} \]

Find Critical Points for Function (d)

Set \( y' = \frac{2 - 8x}{(1-2x)^2} = 0 \), solving gives \( 2 - 8x = 0 \) leading to \( x = \frac{1}{4} \).

Determine Second Derivative for Concavity (d)

Compute the second derivative using the quotient rule again: \[ y'' = \frac{d}{dx}\left(\frac{2-8x}{(1-2x)^2}\right) \] Simplify the expression and evaluate at \( x = \frac{1}{4} \).

Apply the Second-Derivative Test (d)

After simplification, find that \[ y'' \] at \( x = \frac{1}{4} \) is negative, indicating that it is a point of relative maximum.

Key concepts

Critical Points

Critical points in calculus are where the first derivative of a function is zero or undefined. Identifying these points helps us find where a function might change direction—that is, where relative maxima or minima could occur.

In the context of our examples, we set the first derivatives of each function to zero and solve for the variable:

• For function (a) with first derivative \( y' = -4x + 8 \), solving \( -4x + 8 = 0 \) gave us \( x = 2 \).
• For function (b) with \( y' = 3x^2 + 12x \), we factored the equation to find critical points \( x = 0 \) and \( x = -4 \).
• Similarly, for function (c), solving \( x^2 - 6x + 5 = 0 \) resulted in critical points at \( x = 1 \) and \( x = 5 \).
• Lastly, for function (d) \( y' = \frac{2 - 8x}{(1-2x)^2} \), setting \( 2 - 8x = 0 \) gave us \( x = \frac{1}{4} \).

These points are essential as they mark the locations where we need to examine the behavior of the function more closely.

Relative Maximum

A relative maximum, or local maximum, occurs at a point where the function changes direction from increasing to decreasing. This point appears as a peak or high point in a graph.

To identify a relative maximum using the Second-Derivative Test, we evaluate the second derivative at the critical points:

• If \( y'' < 0 \), it indicates concave down, suggesting a relative maximum.

For our exercises:

• In function (a), since \( y'' = -4 \) (concave down), the critical point \( x = 2 \) is a relative maximum.
• For function (b), at \( x = -4 \), \( y'' = -12 \), indicating a relative maximum.
• Similarly, in function (c), at \( x = 1 \), \( y'' = -4 \) also reveals a relative maximum.
• For function (d), the critical point \( x = \frac{1}{4} \) had a second derivative indicating a maximum.

Relative Minimum

A relative minimum occurs where the function changes from decreasing to increasing, forming a trough or low point on the graph. Understanding this concept is crucial as it helps us determine the optimal low points in various contexts.

Using the Second-Derivative Test, relative minima are identified when the second derivative test returns:

• If \( y'' > 0 \), it indicates concave up, suggesting a local minimum.

In our example problems:

• For function (b), the critical point \( x = 0 \) has \( y'' = 12 \), confirming a relative minimum.
• Similarly, in function (c), for \( x = 5 \), \( y'' = 4 \) indicates another relative minimum.

These local minima are helpful for identifying valuable information like cost reduction in business or minimum energy states in physics.

Concavity

Concavity describes how a curve bends and is extremely useful when analyzing the nature of functions and their graphs. It's determined by the second derivative of the function.

• If \( y'' > 0 \), the function is concave up, resembling a "U" shape.
• If \( y'' < 0 \), the function is concave down, resembling an "n" shape.

For the examined functions:

• Function (a) has \( y'' = -4 \), indicating it's concave down at any point, impacting the shape near critical points.
• Function (b) at \( x = 0 \) shows \( y'' = 12 \), thus concave up, while at \( x = -4 \) \( y'' = -12 \), as concave down, clearly describing the function's behavior.
• Function (c) illustrates mixed concavity, demonstrating both concave down at \( x = 1 \) and concave up at \( x = 5 \).
• Function (d) analyzed simple concavity behavior, assisting in geometric interpretation.

Understanding concavity assists in overall comprehension of the function's global shape and behavior, assisting predictions and graphical sketching.