Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Identify and classify all critical points of the function \(f(x)=\left(x^{2}-4\right) x^{\pi+1}\) for \(x>0\).

Step-by-Step Solution

Answer

The function \(f(x)=\left(x^{2}-4\right) x^{\pi+1}\) for \(x>0\) has one critical point at \(x=2\). The classification of this critical point (whether it's a local max or min) depends on the analysis of the second derivative at \(x=2\).

Step 1: Find the derivative

Apply the product rule to compute the derivative of \(f(x)\): \[f'(x) = (2x)x^{\pi+1} + \left(x^{2}-4\right) (\pi+1)x^{\pi}\]

Step 2: Find critical points

Set the derivative equal to zero and solve for \(x\): \[(2x)x^{\pi+1} + \left(x^{2}-4\right) (\pi+1)x^{\pi} = 0.\] Factor out \(x^{\pi}\) and rearrange to get: \[x (2x^{\pi} + (\pi+1)(x^{2}-4)) = 0.\] This gives critical points at \(x=0\) and \(x=2\), \(x=-2.\] However, since \(x>0\), the only relevant critical point is \(x=2\).

Step 3: Find the second derivative

Compute the second derivative \(f''(x)\) to find the concavity of \(f(x)\) at \(x=2\). Apply the product rule and the chain rule, which would involve derivative of \(x^{\pi+1}\) and \(x^{\pi}\), and simplify.

Step 4: Classify the critical point

Substitute \(x=2\) into \(f''(x)\). If \(f''(2)>0\), then \(x=2\) is a local minimum point. If \(f''(2)<0\), then \(x=2\) is a local maximum point. Given the nature of the function, it's most likely a local minimum. However, to make a definite conclusion, plug in \(x=2\) into the second derivative and analyze the result.

Key Concepts

Derivative

The derivative of a function is a fundamental concept in calculus. It represents the rate at which the function's output changes with respect to changes in the input. Think of it as the slope of the tangent line to the function's graph at any point. Calculating a derivative involves finding an expression that tells us how the function behaves. The key to finding derivatives is applying rules such as the power rule, product rule, and chain rule. These rules help simplify the process when dealing with complex functions. In our exercise, the product rule is particularly important because the function involves a product of terms.

Product Rule

The product rule is used to find the derivative of a product of two functions. If you have two functions, say \(u(x)\) and \(v(x)\), and you want to find the derivative of their product \(u(x) v(x)\), you would apply the product rule, which is:

• \((uv)' = u'v + uv'\)

In our exercise, the function \(f(x) = (x^2 - 4) x^{\pi+1}\) is a product of \(x^2 - 4\) and \(x^{\pi+1}\). Using the product rule:

• First, differentiate \(x^2 - 4\) to get \(2x\).
• Then, differentiate \(x^{\pi+1}\) to yield \((\pi+1) x^{\pi}\).

Putting it together, we get the derivative formula for our function. This approach simplifies the process of finding critical points and further analyses.

Concavity

Concavity refers to the curvature of a function's graph. It tells us if the graph is curving upwards (like a cup) or downwards (like a dome). Determining concavity involves the second derivative of the function.

• If the second derivative \(f''(x) > 0\), the function is concave up at that interval. Imagine a happy face curve.
• If \(f''(x) < 0\), the function is concave down. Think of a sad face curve.

In our problem, determining concavity at a critical point helps us classify the point as a local minimum or maximum. If \(f''(x)\) is positive at the critical point, the function is at a local minimum there.

Second Derivative

The second derivative of a function is the derivative of its first derivative. It provides a deeper understanding of the function’s behavior by showing how the rate of change itself is changing. In practical terms, this helps to analyze the concavity of the curve. To find the second derivative, apply the differentiation rules again to \(f'(x)\). In the context of our exercise, you again use the product rule, compounded by differentiation rules like the chain rule, particularly because of terms like \(x^{\pi+1}\) and \(x^{\pi}\). Once the second derivative is determined, evaluate it at the critical point found earlier. This value will confirm whether the critical point is a point of minima or maxima, aiding in classifying the critical point.