Problem 26 - Calculus: Early Transcendentals

Question

Find all local maximum and minimum points by the second derivative test. $$ y=(x+1) / \sqrt{5 x^{2}+35} $$

Step-by-Step Solution

Answer

Local maximum at x = 7, y(7) = 4/√70.

Step 1: Find the first derivative

\(y = \frac{x+1}{\sqrt{5x^2+35}}\). Using the quotient rule:
\(y' = \frac{\sqrt{5x^2+35} - (x+1) \cdot \frac{5x}{\sqrt{5x^2+35}}}{5x^2+35} = \frac{5x^2+35-5x^2-5x}{(5x^2+35)^{3/2}} = \frac{5(7-x)}{(5x^2+35)^{3/2}}\)

Step 2: Find critical points and classify

\(y' = 0\) when \(7 - x = 0\), i.e., \(x = 7\).

For \(x < 7\): \(y' > 0\) (increasing). For \(x > 7\): \(y' < 0\) (decreasing). So \(x = 7\) is a local maximum by the first derivative test.

\(y(7) = \frac{8}{\sqrt{5(49)+35}} = \frac{8}{\sqrt{280}} = \frac{8}{2\sqrt{70}} = \frac{4}{\sqrt{70}}\)

Key Concepts

Quotient Rule

The quotient rule is a useful technique for finding the derivative of a function that is the ratio of two differentiable functions. It comes in handy when you need to handle divisions in calculus. When you have a function like \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule is deployed.

• The formula is: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). This means you take the derivative of the numerator multiplied by the denominator, minus the numerator multiplied by the derivative of the denominator, and divide the whole thing by the square of the denominator.
• Important: This rule prevents mistakes that could occur by trying to differentiate the fraction as a whole without this formula.

In our original exercise, you need to please pay attention while substituting the values for \( u \), \( u' \), \( v \), and \( v' \). Each mistake in these calculations could lead to wrong results in your derivatives.

First Derivative

The first derivative of a function reveals the rate at which the function's value is changing. This is crucial as it tells us about the slope of the function at any given point on its curve.

• Calculating the first derivative is often the first step in finding critical points, which are candidates for local maxima and minima.
• For the function \( y = \frac{x+1}{\sqrt{5x^2 + 35}} \), the first derivative was found using the quotient rule, resulting in a somewhat complex expression that needs simplification before further steps.

Recognizing and accurately computing the first derivative ensures that you move forward correctly to the steps of finding and confirming critical points.

Local Maximum and Minimum

Local maximum and minimum points are points on a graph where the function reaches its highest or lowest value within a neighboring interval.

• We use the first derivative to locate these points by setting it to zero to find critical points.
• After calculating critical points, the second derivative test can confirm whether these are maxima, minima, or points of inflection.

If the second derivative is positive at a critical point, the function has a local minimum there. If it's negative, the function has a local maximum. An easy-to-understand way to remember this is to think: a smile (upward parabola) is positive and a frown (downward parabola) is negative.

Calculus Problem Solving

Solving calculus problems often involves a series of steps that must be performed methodically to ensure accuracy and understanding. Getting to the solution of finding local maxima and minima with derivatives involves:

• Understanding which rules and formulas to apply, such as the quotient rule in our exercise.
• Clearly breaking down the steps as you calculate the first derivative, and then the second if necessary.
• Using tests like the second derivative test to conclude about the nature of critical points.

Problem-solving in calculus is not just about finding an answer but understanding the process along the way. Each step builds into the next, emphasizing the importance of thorough knowledge of derivatives and their implications in understanding function behavior.