Question

Show using calculus that \(x_{0}=\frac{2 b}{3}\) does maximize the function \(x^{2}(b-x)\). Then use calculus to analyze the graph of \(y=x^{3}-b x^{2}+d\) and confirm Sharaf al-Din's conclusion on the number of positive solutions to \(x^{3}+d=b x^{2}\).

Short answer

Based on the steps provided, the conclusion is that there is only one positive solution to the equation \(x^3 + d = bx^2\), which happens at \(x = \frac{2b}{3}\). This confirms Sharaf al-Din's claim and shows that the given point \(x_0=\frac{2b}{3}\) maximizes the function \(x^2(b-x)\).

Step-by-step solution

Find the first derivative of \(x^2(b-x)\)

To find the first derivative, we will apply the product rule on the function \(x^2(b-x)\): \[ \frac{d}{dx}(x^2(b-x)) = \frac{d}{dx}(x^2) * (b-x) + x^2 * \frac{d}{dx}(b-x). \] Now, differentiate each component in the equation: \[ \frac{d}{dx}(x^2(b-x)) = 2x(b-x) - x^2. \]

Set the first derivative equal to zero to find critical points

To find the critical points of the function, we need to find the values of \(x\) for which the first derivative is equal to zero: \[ 2x(b-x) - x^2 = 0. \] Factor \(x\) out of the equation: \[ x(2b - 3x) = 0. \] We obtain two critical points, \(x=0\) and \(x=\frac{2b}{3}\).

Find the second derivative of \(x^2(b-x)\)

Now, we will take the first derivative and differentiate it once more to obtain the second derivative: \[ \frac{d^2}{dx^2}(x^2(b-x)) = \frac{d}{dx}(2x(b-x) - x^2). \] Differentiate the first derivative: \[ \frac{d^2}{dx^2}(x^2(b-x)) = 2(b-x) - 2x - 2x = -6x + 2b. \]

Analyze the critical points using the second derivative

We will use the second derivative test to determine the nature of the critical points. If the second derivative is positive, the critical point is a local minimum, while if the second derivative is negative, the critical point is a local maximum. At \(x=0\): \[ -6(0) + 2b = 2b. \] Since \(b>0\), at \(x=0\), the second derivative is positive, indicating a local minimum. At \(x=\frac{2b}{3}\): \[ -6 \left(\frac{2b}{3}\right) + 2b = -4b. \] Since \(b>0\), at \(x=\frac{2b}{3}\), the second derivative is negative, indicating a local maximum. Thus, we have shown that \(x_0=\frac{2b}{3}\) maximizes the function \(x^2(b-x)\).

Find critical points of \(y=x^3-bx^2+d\)

Now, let's analyze the graph of \(y=x^3-bx^2+d\). First, we'll find the critical points by finding the first derivative and set it equal to zero. \[ \frac{d}{dx}(x^3-bx^2+d) = 3x^2 - 2bx. \] Setting this equal to zero, we get: \[ 3x^2 - 2bx = 0. \] Factor out \(x\): \[ x(3x - 2b) = 0. \] We obtain two critical points \(x=0\) and \(x=\frac{2b}{3}\).

Use critical points to confirm Sharaf al-Din's conclusion

We found two critical points \(x=0\) and \(x=\frac{2b}{3}\) for both functions. This confirms that there is only one positive solution to \(x^3 + d = bx^2\) as claimed by Sharaf al-Din, which occurs at \(x = \frac{2b}{3}\).

Key concepts

Critical Points in Calculus

Understanding critical points is fundamental in calculus, as they reveal where a graph may change direction, increase, or decrease. Critical points occur where the first derivative of a function either equals zero or does not exist.

To find these points, you begin by taking the derivative of the function with respect to its variable, which, in our exercise, is symbolized as \( \frac{d}{dx}(x^2(b-x)) \). The critical points are then found by setting this derivative equal to zero and solving for the variable, here represented by \( x \).

In practical terms, critical points can help determine the maximum and minimum values of a function on a certain interval. For instance, the critical points for \( x^2(b-x) \) led us to discover that \( x_0=\frac{2b}{3} \) is a local maximum – a point of great interest when analyzing the behavior of functions graphically or in applied contexts.

Second Derivative Test

The second derivative test is a convenient method to determine the nature of a critical point found on a function. After finding the first derivative and setting it to zero to locate critical points, you take the second derivative of the function.

The sign of the second derivative at a critical point will tell you whether that point is a local minimum (if the second derivative is positive) or a local maximum (if the second derivative is negative).

• If \( \frac{d^2}{dx^2} > 0 \) at the critical point, we have a local minimum.
• If \( \frac{d^2}{dx^2} < 0 \) at the critical point, we have a local maximum.

In the provided exercise, we applied this test and found that for \( x_0=\frac{2b}{3} \) the function \( x^2(b-x) \) has indeed a local maximum, as the second derivative at this point is negative. This step is invaluable for confirming the nature of critical points without the need to graphically plot the function.

Sharaf al-Din's Theorem

Sharaf al-Din al-Tusi made significant contributions to algebra and calculus, including a theorem on cubic equations like \( x^3 + d = bx^2 \). Analyzing how the understanding of critical points can lead to graphical interpretation of functions, the theorem posits conditions under which specific types of cubic equations will have positive solutions.

To delve into Sharaf al-Din's findings, we leveraged calculus to analyze the function \( y=x^3-bx^2+d \) by finding its critical points. In our exercise, the correspondence of critical points between \( x^2(b-x) \) and \( y=x^3-bx^2+d \) confirms the theorem's claim by showing there's only one positive solution to the equation.

This intersection of historical mathematics and modern calculus serves as a reminder of the depth and relevance of classical works in our current mathematical endeavours. It can enhance student appreciation for the evolution of mathematical thought and its applications.