Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find and classify all critical points. $$ f(x)=x^{3}+2 x^{2}+3 x+4 $$

Step-by-Step Solution

Answer

The function \(f(x)=x^{3}+2 x^{2}+3 x+4\) has one critical point at \(x=-1\), which is a local maximum.

Step 1: Compute the Derivative

The first step is to compute the derivative of the function \(f(x)\). The derivative of \(f(x)=x^{3}+2 x^{2}+3 x+4\) is \(f'(x)=3x^{2}+4x+3\).

Step 2: Find Critical Points

Critical points occur where the derivative of the function is equal to zero or undefined. However, since \(f'(x)\) is a polynomial, it's never undefined. So set \(3x^{2}+4x+3=0\) and solve for \(x\). This is a quadratic equation, and we can solve it using the quadratic formula \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). In this case, \(a=3, b=4\) and \(c=3\). Solving this gives \(x=-1, -1\), two identical real roots.

Step 3: Compute the Second Derivative

Now that the critical points are known, it is time to classify them. Compute the second derivative of the function. The second derivative of \(f(x)\) is \(f''(x)=6x+4\).

Step 4: Classify the Critical Points

Substitute the critical points into the second derivative. \(f''(-1)=6(-1)+4=-2\). Because \(f''(-1)<0\), the second derivative test tells us that \(x=-1\) is a local maximum.

Key Concepts

Derivative

When we talk about derivatives in calculus, we refer to a powerful tool that helps understand how functions change. A derivative essentially measures the rate at which a function's value changes as its input changes. For any function \(f(x)\), finding its derivative \(f'(x)\) involves using differentiation rules. Here, we address a particular polynomial function:

• Given \(f(x) = x^3 + 2x^2 + 3x + 4\), the derivative is \(f'(x) = 3x^2 + 4x + 3\).
• This result is obtained by applying the power rule: bring down the exponent as a coefficient, and then decrease the exponent by one.

Derivatives are crucial for analyzing critical points, which reveal where a function's graph has potential turning points. It’s important to know that in a polynomial like this, the derivative will never be undefined, so we purely focus on where it equals zero.

Polynomial Function

Polynomials are a central topic in algebra and calculus, made up of terms consisting of variables raised to non-negative integer powers. The polynomial in this exercise, \(f(x) = x^3 + 2x^2 + 3x + 4\), is a cubic function.

• The highest power, or degree, is 3, indicating it’s a cubic polynomial.
• The shape of cubic polynomials can show multiple turns in their graphs, with peaks and valleys, known as local maxima and minima.
• As polynomials, they are continuous and smooth across their domain, without any breaks or gaps.

The quadratic equation \(3x^2 + 4x + 3 = 0\) is derived when seeking the critical points to set the derivative to zero. Solving this quadratic tells us where these points, or turning points, might be located. Through the quadratic formula, we find the roots: \(x = -1\), indicating the symmetry often seen in polynomial roots.

Second Derivative Test

The second derivative test is a method used to classify critical points found from the first derivative. By computing the second derivative \(f''(x)\) and evaluating it at critical points, one can determine the nature of these points.

• For \(f(x) = x^3 + 2x^2 + 3x + 4\), the second derivative comes out to \(f''(x) = 6x + 4\).
• This test classifies points based on the sign of the second derivative:
• If \(f''(x) > 0\), the function has a local minimum at that point.
• If \(f''(x) < 0\), there's a local maximum.
• If \(f''(x) = 0\), the test is inconclusive.

Applying this to the critical point \(x = -1\), we find \(f''(-1) = -2\), which is less than zero, thus confirming a local maximum at that spot. This test adds depth to understanding how the function behaves around each critical point, revealing the full picture of its changing nature.