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Magazine Chapter 5: Problem 15
Find all local maximum and minimum points by the second derivative test, when possible. $$y=(x+5)^{1 / 4}$$
Short Answer
Expert verified
No local maxima or minima exist due to the domain constraint at \( x = -5 \).
Step by step solution
01
Find the First Derivative
To find the first derivative of the function \( y = (x+5)^{1/4} \), use the power rule. The power rule states that if \( y = x^n \), then \( y' = nx^{n-1} \). Here, \( n = \frac{1}{4} \). Compute the derivative: \( y' = \frac{1}{4}(x+5)^{\frac{1}{4} - 1} = \frac{1}{4}(x+5)^{-3/4} \).
02
Find the Critical Points
Critical points occur where the first derivative is equal to zero or undefined. Since \( y' = \frac{1}{4}(x+5)^{-3/4} \) will never be zero, solve \( \frac{1}{4}(x+5)^{-3/4} = 0 \) to detect points where it is undefined. As \((x+5)^{-3/4}\) becomes undefined when \( x+5=0 \), conclude that the critical point is \( x = -5 \).
03
Calculate the Second Derivative
The function \( y' = \frac{1}{4}(x+5)^{-3/4} \) is now differentiated again to find the second derivative \( y'' \). Using the power rule for differentiation: \( y'' = -\frac{3}{16}(x+5)^{-7/4} \).
04
Apply the Second Derivative Test
The second derivative test involves substituting the critical point into the second derivative. For \( x = -5 \), \( y''(-5) \) is undefined because the expression \( (x+5)^{-7/4} \) is undefined at \( x = -5 \). Thus, the second derivative test is inconclusive here.
05
Check for Local Extrema
Since the second derivative test is inconclusive at \( x = -5 \), investigate the first derivative \( y' = \frac{1}{4}(x+5)^{-3/4} \). As \( x = -5 \) is the domain boundary, the behavior of \( y \) as \( x \) approaches \( -5 \) is the best indication of potential extrema. Since the function \( y \) is increasing for \( x > -5 \) and undefined for \( x < -5 \), we conclude there are no traditional local maxima or minima.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Critical points are vital in calculus as they help determine where a function might have a high or low point, such as a local maximum or minimum. To find these points, the key is to examine the first derivative of the function:
- A critical point occurs where the first derivative is zero. However, it can also occur where the first derivative is undefined.
- In our example, with the function \( y = (x+5)^{1/4} \), the first derivative \( y' = \frac{1}{4}(x+5)^{-3/4} \) never reaches zero. However, it's undefined when \( x + 5 = 0 \), leading to the critical point at \( x = -5 \).
First Derivative
The first derivative of a function essentially represents the slope or rate of change at any given point along the function's curve. It provides crucial insights into how the function behaves:
- It tells us if the function is increasing or decreasing at a particular point. Positive values indicate the function is going upwards, while negative values mean it's declining.
- By using the power rule, the first derivative for \( y = (x+5)^{1/4} \) is computed as \( y' = \frac{1}{4}(x+5)^{-3/4} \). This derivative helps identify critical points where changes such as turning from increasing to decreasing can occur.
Local Maxima and Minima
Local maxima and minima are points where a function reaches a local high or low, meaning the function's value doesn't exceed this point in its immediate vicinity. Two key steps help identify these points:
- First, find the critical points using the first derivative.
- Then, apply the second derivative test to distinguish between a local maximum, minimum, or if the test result is inconclusive.
Calculus
Calculus is a rich field of mathematics focused on studying change and motion. It's mainly divided into two branches: Differential and Integral Calculus.
- Differential Calculus examines the rate of change and slopes, primarily through derivatives like the first and second derivatives we explored in our function \( y = (x+5)^{1/4} \).
- Integral Calculus focuses on summation and areas under curves. Although not directly dealt within this problem, it's equally important in broader calculus studies.
