Problem 41 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the maximum value of \(\left|f^{\prime \prime}(x)\right|\) on the closed interval. (You will use this skill in Section \(12.4\) to estimate the error in the Trapezoidal Rule.) $$ f(x)=\sqrt{1+x^{3}}, \quad[0,2] $$

Step-by-Step Solution

Answer

The maximum value of \(|f''(x)|\) on [0,2] is either the maximum value of \(|f''(x)|\) at the critical point(s), 0 or 2.

Step 1: Finding the First Derivative

Finding the derivative of a function involves applying the rules of differentiation. In this case, the Chain Rule is required to differentiate the function. The Chain Rule involves differentiating the outer function and then the inner function. The derivative of the outer function \(\sqrt{x}\) is \(0.5x^{-0.5}\) and the inner function \(1 + x^{3}\) is \(3x^{2}\). Therefore, the derivative \(f'(x)\) of the function \(f(x)=\sqrt{1+x^{3}}\) is: \(f'(x)= \frac {1}{2}(1+x^{3})^{-1/2} * 3x^{2}\).

Step 2: Finding the Second Derivative

The next step is to find the second derivative, \(f''(x)\), of the function. This can be done by differentiating the first derivative \(f'(x)\) using the Product Rule and Chain Rule together. It is easier to write \(f'(x)\) as \( \frac {3x^{2}}{2}\times(1+x^{3})^{-1/2}\). Now, the Product Rule can be applied which states that the derivative of \(u(x)\times v(x)\) is \(u'(x)v(x)+u(x)v'(x)\). Using the Chain Rule for the derivative of \(v(x)=(1+x^{3})^{-1/2}\) will yield \(v'(x)= -1/2 (1+x^{3})^{-3/2} * 3x^{2}\). Therefore, \(f''(x)= [3/2 * \frac {d}{dx} (1+x^{3})^{-1/2}]\times(1+x^{3})^{-1/2} + \frac {3x^{2}}{2} \times [-1/2 (1+x^{3})^{-3/2} * 3x^{2}]. Simplify \(f''(x)\) to get the final form of the second derivative.

Step 3: Finding the Maximum Value of |f''(x)| on [0,2]

Finally, to find the maximum value of \(|f''(x)|\), the critical points within the interval [0,2] need to be found. This is done by setting the second derivative equal to zero and solving for \(x\). It is important to also evaluate \(f''(x)\) at the boundary points of the interval which are 0 and 2. The greater absolute value among all these evaluations is the maximum value of \(|f''(x)|\) on [0,2].

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus, especially useful for differentiating composite functions. When you have a function nested within another function, like in the problem where we have \( f(x) = \sqrt{1 + x^3} \), the Chain Rule helps us find the derivative.
For a function \( g(f(x)) \), the derivative using the Chain Rule is given by \( g'(f(x)) \cdot f'(x) \).

In simpler terms, differentiate the outer function and multiply by the derivative of the inner function.

• The outer function here is \( \sqrt{x} \), which has the derivative \( 0.5x^{-0.5} \).
• The inner function is \( 1 + x^3 \), with the derivative \( 3x^2 \).

By applying the Chain Rule, we find the first derivative: \( f'(x) = \frac{1}{2}(1+x^3)^{-1/2} \cdot 3x^2 \). This skill is invaluable when tackling more complex functions in calculus.

Product Rule

In calculus, when you take the derivative of a product of two functions, the Product Rule comes into play. The Product Rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product is given by \( u'(x)v(x) + u(x)v'(x) \).

This is crucial when computing higher-order derivatives, such as the second derivative \( f''(x) \), which appears in our original problem. When \( f'(x) \) is written as a product \( \frac{3x^2}{2} \cdot (1 + x^3)^{-1/2} \), you should apply the Product Rule.

• Here, let \( u(x) = \frac{3x^2}{2} \) with \( u'(x) = 3x \).
• Let \( v(x) = (1 + x^3)^{-1/2} \). Use the Chain Rule to find \( v'(x) = -\frac{1}{2}(1 + x^3)^{-3/2} \cdot 3x^2 \).

Using these derivatives, the Product Rule helps calculate \( f''(x) \), an essential step in solving our problem.

Second Derivative

The second derivative of a function is crucial for understanding the behavior of the function, especially when looking for points of inflection or concavity. In our exercise, finding \( f''(x) \) involves differentiating the first derivative we obtained with techniques like the Chain Rule and Product Rule.

• The second derivative, \( f''(x) \), involves applying both rules to streamline the process and eventually simplifying the result for easier computation.
• In our case, computation would yield \( f''(x) \) after simplification. This derivative will then be used to find maximum values of its absolute form in a given interval.

The second derivative gives us greater insight into the changes and extremes, like acceleration in physics contexts or curvatures in geometric considerations.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points signify where important changes happen in the behavior of the function. To find where \( |f''(x)| \) is maximized in a specific interval, finding and evaluating critical points is essential.

• For our problem, critical points arise from the second derivative \( f''(x) = 0 \).
• We must also check the endpoints of the interval [0, 2] to determine where the greatest absolute change occurs.

By checking these points, you can identify the maximum value of \( |f''(x)| \). This value provides insights, for example, into the range of deviations when employing numerical integration methods such as the Trapezoidal Rule. Understanding critical points not only helps in calculus but has practical applications in optimization problems across various fields.