Problem 32 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the point(s) of inflection of the graph of the function. $$ f(x)=x(6-x)^{2} $$

Step-by-Step Solution

Answer

The point of inflection of the graph of the function \(f(x) = x(6-x)^2\) is at (3,27)

Step 1: Find the first derivative

To find the first derivative, apply the product rule which states that the derivative of two functions multiplied together is the first function times the derivative of the second plus the second function times the derivative of the first. Using this rule, the derivative of \(f(x) = x(6-x)^{2}\) would be \(f'(x) = (6-x)^{2} - 2x(6-x)\). Simplifying this gives \(f'(x)= -2x^{2} + 12x - 36\)

Step 2: Find the second derivative

Differentiate the result from the first step again to get the second derivative \(f''(x)\). This is straightforward as it only involves the power rule. The second derivative is \(f''(x) = -4x + 12\)

Step 3: Solve \(f''(x) = 0\)

Find the inflection point, set \(f''(x) = 0\) and solve for \(x\). This gives \(x = 3\)

Step 4: Find the y-coordinate

Plug in \(x = 3\) into the original function to get the corresponding y-value. Use \(f(x) = x(6-x)^2\) to give \(f(3) = 3(6-3)^2 = 27\)

Step 5: Identify the point of inflection

Thus, the point of inflection of the graph of the function is (3,27)

Key Concepts

Understanding Derivatives

A derivative represents how a function changes at any point on its graph. It gives us the rate at which the function's value is changing with respect to its variable. Think of it as the slope of the tangent line at a particular point.

In our exercise, the function is \( f(x) = x(6-x)^2 \). To find the derivative, we need to understand how each part of this function changes. This helps us find where the graph's shape changes significantly, such as where it curves upwards or downwards.

Derivatives are essential in identifying critical points, such as maxima, minima, and inflection points, by helping us see where the growth pattern of a function changes.

Product Rule in Action

The product rule is crucial when differentiating functions that are products of two smaller functions. If you have a function \( f(x) = u(x) \cdot v(x) \), the product rule states that the derivative \( f'(x) \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). This helps us break down and handle complex functions by focusing on each component separately.

In this exercise, we apply the product rule to \( f(x) = x(6-x)^2 \). Here:

• \( u(x) = x \)
• \( v(x) = (6-x)^2 \)

By applying the product rule, we derive \( f'(x) = (6-x)^2 - 2x(6-x) \). It allows us to capture how each part of the product contributes to the overall rate of change.

Applying the Power Rule

The power rule simplifies taking derivatives of polynomial functions. It's a straightforward rule: if \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \). This straightforward method makes finding derivatives quick and efficient.

In this problem, after using the product rule, we needed to find the second derivative \( f''(x) \) of the simplified expression \( -2x^2 + 12x - 36 \).

By applying the power rule:

• The derivative of \( -2x^2 \) is \( -4x \)
• The derivative of \( 12x \) is \( 12 \)
• The constant \( -36 \) becomes \( 0 \)

Hence, \( f''(x) = -4x + 12 \). This second derivative is essential for finding inflection points, where the function's concavity changes.