Problem 31 - Calculus: Early Transcendentals

Question

Find all local maximum and minimum points by the second derivative test. $$ y=(x+5)^{1 / 4} $$

Step-by-Step Solution

Answer

There are no local maximum or minimum points because the function has no critical points.

Step 1: Find the First Derivative

To find the critical points, start by finding the first derivative of the function. The function given is \( y = (x+5)^{1/4} \). Using the power rule for differentiation, the first derivative is \( y' = \frac{1}{4}(x+5)^{-3/4} \).

Step 2: Find Critical Points

Set the first derivative equal to zero to find critical points. Solve \( \frac{1}{4}(x+5)^{-3/4} = 0 \). There are no real solutions since a non-zero power cannot equal zero. Check where the derivative is undefined: it's undefined at \( x = -5 \). However, since the function itself is not defined for \( x < -5 \), we don't include \( x = -5 \) as a critical point.

Step 3: Find the Second Derivative

Find the second derivative of the original function to apply the second derivative test. Differentiate \( y' = \frac{1}{4}(x+5)^{-3/4} \) to get \( y'' = -\frac{3}{16}(x+5)^{-7/4} \).

Step 4: Apply the Second Derivative Test

Since there are no critical points from where \( y' = 0 \), check if the function can have behavior changes anywhere else inside its domain. Evaluate the second derivative at a point in its domain, for instance, \( x = 0 \). At \( x = 0 \), \( y'' = -\frac{3}{16}(5)^{-7/4} \) which is negative, implying the function is concave down, but without a critical point, there is no local extremum.

Key Concepts

First Derivative

The first derivative of a function is essential for understanding its behavior and rates of change. For the function given, which is \( y = (x+5)^{1/4} \), the first derivative helps us establish the slope at any given point. To find it, we utilize the power rule in calculus. The power rule states that if you have a term \( x^n \), its derivative is \( nx^{n-1} \). Applying this rule to our function gives us:

• \( y' = \frac{1}{4}(x+5)^{-3/4} \)

This derivative tells us how \( y \) changes as \( x \) changes. It is crucial in determining where slopes are zero or where they don't exist, which are the cases where critical points can occur.

Critical Points

Critical points are values of \( x \) where the derivative is zero or undefined. These points are vital as they indicate where a function can have potential maxima, minima, or points of inflection. To find critical points for \( y = (x+5)^{1/4} \), we set the first derivative \( y' \) to zero and solve for \( x \). In this situation:

• \( \frac{1}{4}(x+5)^{-3/4} = 0 \)

However, there are no real solutions since a term raised to a power won't equal zero unless that term can be zero itself, which isn't the case here. Next, check where the derivative is undefined: \( x = -5 \). But, since our function is undefined for \( x < -5 \), we exclude this as a critical point. Therefore, this exploration reveals no critical points in \( (x+5)^{1/4} \), meaning the second derivative test cannot find local extrema based on critical points.

Concavity

Concavity describes the direction a function curves. Understanding concavity involves finding the second derivative of the function. By doing this, we learn whether the function is bending upwards or downwards. Calculating the second derivative of our function:

• \( y'' = -\frac{3}{16}(x+5)^{-7/4} \)

Checking the concavity at a specific point in the function's domain can give insight into the function's shape. For example, at \( x = 0 \):

• \( y'' = -\frac{3}{16}(5)^{-7/4} \)

This negative result implies the function is concave down at this point in its domain. Points of concavity change, often occurring at critical points where second derivative tests help determine local maxima or minima. In this case, without critical points from \( y' = 0 \), the analysis indicates no local extrema based on concavity.

Power Rule

The power rule is a fundamental principle in calculus used to differentiate functions of the form \( x^n \). It simplifies the process of finding derivatives by following a straightforward pattern: multiply the function by the power and decrease the power by one. In our exercise with \( y = (x+5)^{1/4} \), the power rule allows us to efficiently compute the first derivative:

• If \( y = (x+5)^{1/4} \), then \( y' = \frac{1}{4}(x+5)^{-3/4} \)

Recognizing when to apply the power rule is crucial for solving many types of calculus problems. It's particularly useful when you need to find both first and second derivatives, as each successive differentiation involves further applications of this rule. Understanding this rule not only aids in solving derivative problems but also builds foundational skills necessary for more advanced calculus topics.