Problem 23 - Applied Calculus

Question

(a) Find all critical points and all inflection points of the function \(f(x)=x^{4}-2 a x^{2}+b .\) Assume \(a\) and \(b\) are positive constants. (b) Find values of the parameters \(a\) and \(b\) if \(f\) has a critical point at the point (2,5) (c) If there is a critical point at \((2,5),\) where are the inflection points?

Step-by-Step Solution

Answer

Critical points at \(x = 0, \pm 2\); Inflection points at \(x = \pm \frac{2\sqrt{3}}{3}\). \(a=4, b=5\).

Step 1: Find the First Derivative

The first step is to find the first derivative of the function \(f(x)=x^4 - 2ax^2 + b\). Differentiating term by term, we get:\[ f'(x) = 4x^3 - 4ax.\]

Step 2: Critical Points Calculation

Critical points occur where the first derivative is zero or undefined. Here, set the first derivative equal to zero:\[ 4x^3 - 4ax = 0. \]Factor out the common terms:\[ 4x(x^2 - a) = 0. \]The solutions are \(x = 0\) and \(x = \pm\sqrt{a}.\) Thus, the critical points in terms of \(a\) are at \(x = 0, \sqrt{a}, -\sqrt{a}.\)

Step 3: Second Derivative for Inflection Points

Next, find the second derivative to determine the inflection points. Differentiating the first derivative:\[ f''(x) = 12x^2 - 4a. \]Set the second derivative equal to zero:\[ 12x^2 - 4a = 0 \]Solve for \(x\):\[ x^2 = \frac{a}{3}. \]Thus, the inflection points occur at \(x = \pm\sqrt{\frac{a}{3}}.\)

Step 4: Determine Parameter Values

We know there is a critical point at \((2,5)\). For this, substitute \(x = 2\) into \(f'(x) = 0\):\[ 4 \times 2^3 - 4a \times 2 = 0. \]Solve for \(a\):\[ 32 - 8a = 0 \rightarrow a = 4. \]Now, substitute \((2,5)\) into the original function to find \(b\):\[ f(2) = 2^4 - 2 \times 4 \times 2^2 + b = 5. \]\[ 16 - 16 + b = 5 \rightarrow b = 5. \]

Step 5: Inflection Points with a and b

With \(a = 4\), calculate the inflection points:\[ x = \pm \sqrt{\frac{4}{3}} = \pm \frac{2\sqrt{3}}{3}. \]

Key Concepts

Inflection Points

Inflection points of a function are specific locations where the curve changes its concavity. This means, the curve will shift from being concave up (curving upwards) to concave down (curving downwards), or vice versa.
To find inflection points, we use the second derivative of the function. An inflection point occurs where the second derivative equals zero, assuming that it changes sign around that point.
For our function, we take the second derivative:

• \( f''(x) = 12x^2 - 4a \).

We solve the equation \( 12x^2 - 4a = 0 \) to find potential inflection points.
The result is \( x = \pm \sqrt{\frac{a}{3}} \). This provides us the points where the function could change concavity. It's crucial to check the second derivative around these points to confirm if they indeed change signs, thereby confirming the presence of inflection points.

First Derivative

The first derivative of a function gives us valuable information about the slope or gradient of the function. It helps in understanding how the function is increasing or decreasing at any given point.
For our function, the first derivative is calculated as:

• \( f'(x) = 4x^3 - 4ax \).

To find critical points, we set this derivative equal to zero. Doing so provides us the points where the slope is zero, often corresponding to maxima, minima, or saddle points. By setting \( 4x^3 - 4ax = 0 \), we find that the critical points are at \( x = 0, \sqrt{a}, -\sqrt{a} \).

• These are potential locations where the function's direction of movement changes.

Second Derivative

The second derivative of a function gives us information about the concavity of the function. A positive second derivative indicates that the function is concave up, whereas a negative second derivative implies the function is concave down.
We derive the second derivative from the first derivative:

• The calculation for our function yields \( f''(x) = 12x^2 - 4a \).

By setting \( f''(x) = 0 \), we find the points where concavity might change, which are potential inflection points described earlier. The second derivative test is crucial in confirming whether these critical points are local minima, maxima, or points of inflection. It provides a deeper insight into the nature of the function's curve.

Calculus

Calculus is a branch of mathematics that studies change. It involves two key concepts - derivatives and integrals. In our context, calculus helps us understand the behavior and characteristics of functions through differentiation.
Differentiation, the process of finding derivatives, allows us to determine:

• Slopes (using the first derivative)
• The rate of change
• Concavity or convexity of functions (using the second derivative)

By applying these calculus techniques, we can gain detailed insights into a function's behavior. Calculus not only assists in solving mathematical problems but also offers a framework to model and analyze real-world situations where change is involved.