Question

In Exercises, find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\). $$ f(x)=(x+2)(x-2)(x+3)(x-3) $$

Short answer

The second derivative of the function \(f(x)=(x+2)(x-2)(x+3)(x-3)\) is \(f''(x)=12x^2\). The equation \(f''(x) = 0\) has a solution at \(x=0\)

Step-by-step solution

State the Given Function

The given function is \(f(x)=(x+2)(x-2)(x+3)(x-3)\).

Find the First Derivative

To find the first derivative, we'll make use of the product rule which states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Applying this rule in the given function implies that: \[f'(x)= (x+2)(x-2)(x+3)\cdot 1+ (x+2)(x-2)(x-3)\cdot 1 + (x+2)(x+2)(x-3)\cdot 1 + (x+2)(x-2)(x+3)\cdot 1\]. Simplifying the above results in: \[f'(x)=4x^3\].

Find the Second Derivative

Now, take the derivative of the first derivative to find the second derivative: \[f''(x) =12x^2\].

Solve the Equation

Now, to find the values of \(x\) where the second derivative equals zero, we set \(f''(x)=0\). Solving the equation \(12x^2=0\) for \(x\) we get \(x=0\) as the solution.

Key concepts

Second Derivative

The second derivative of a function provides important information about the function's concavity and points of inflection. After differentiating a function once to find its first derivative, you differentiate it again to obtain the second derivative. This second derivative tells us how the slope of the tangent line is changing, which in other words, indicates the function's behavior. If the second derivative is positive over an interval, the function is concave up (like a smile). If it's negative, the function is concave down (like a frown). When solving problems involving second derivatives, you might be asked to solve an equation to find critical points, such as in cases where the second derivative equals zero. These critical points can sometimes indicate points of inflection, where the concavity changes direction.

Product Rule

The product rule is a fundamental technique in differentiation used when you need to find the derivative of a product of two or more functions. If you have two functions, say \(u(x)\) and \(v(x)\), the product rule states that the derivative of their product is given by:

• \(\frac{d}{dx}[u(x)\cdot v(x)] = u'(x)\cdot v(x) + u(x)\cdot v'(x)\)

This rule is crucial in problems where a function is expressed as the product of multiple factors. In our exercise, the function parameterized by \((x+2)(x-2)(x+3)(x-3)\) was differentiated using the product rule. At each step, it required careful application to each pair of multiplying factors.

Polynomial Function

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. An example is \(f(x) = x^4 - 13x^2 + 36\). Polynomials are the building blocks for many functions in calculus and are particularly straightforward to work with when differentiating. The degree of the polynomial, determined by the highest power of \(x\), guides the behavior of the graph at both ends and includes finding derivatives. The exercise provided had a fourth-degree polynomial, so when simplified, its structure allowed us to easily apply differentiation rules such as the product rule and find its derivatives by recognizing patterns in terms of degrees and coefficients, such as extracting a cubic derivative from it.

Solving Equations

Solving equations in calculus often involves setting a derivative equal to zero to find critical points. For derivatives, these solutions indicate important values for which the function's behavior changes. For our problem, this involved the second derivative being set to zero:

• \(f''(x) = 12x^2 = 0\)

To solve this, rearrange the equation and solve for \(x\), yielding the solution \(x = 0\). This step is essential for identifying intercepts and using them to determine the points which could be either maxima, minima, or points of inflection, depending on further testing or understanding of the function's behavior over an interval.