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Chapter 2: Problem 38

The graph of each function has one relative extreme point. Find it (giving both \(x\) - and \(y\) -coordinates) and determine if it is a relative maximum or a relative minimum point. Do not include a sketch of the graph of the function. $$ g(x)=x^{2}+10 x+10 $$

Short Answer

Expert verified
The relative minimum point is at \( (-5, -15) \).

Step by step solution

01

- Find the first derivative

To find the relative extreme point of the function, first compute the derivative of the function. For the function \( g(x) = x^2 + 10x + 10 \), the first derivative is \( g'(x) \). Differentiate each term separately:\[ g'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(10x) + \frac{d}{dx}(10) \]Applying the power rule and constant rule:\[ g'(x) = 2x + 10 \]
02

- Set the first derivative to zero

Set the derivative equal to zero and solve for \( x \):\[ 2x + 10 = 0 \]Subtract 10 from both sides:\[ 2x = -10 \]Divide both sides by 2:\[ x = -5 \]
03

- Find the second derivative

To determine if the critical point is a maximum or minimum, find the second derivative of the function. For the function \( g(x) = x^2 + 10x + 10 \), differentiate the first derivative \( g'(x) = 2x + 10 \):\[ g''(x) = \frac{d}{dx}(2x + 10) = 2 \]
04

- Analyze the second derivative

Evaluate the second derivative. Since \( g''(x) = 2 \), which is greater than 0, the critical point at \( x = -5 \) is a relative minimum.
05

- Calculate the y-coordinate

Substitute \( x = -5 \) back into the original function to find the corresponding \( y \)-coordinate:\[ g(-5) = (-5)^2 + 10(-5) + 10 \]Simplify:\[ g(-5) = 25 - 50 + 10 = -15 \]So the coordinates of the relative minimum point are \( (-5, -15) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative
The first derivative of a function, often denoted as \( f'(x) \), provides crucial information about the function's rate of change. For a given function \( g(x) = x^2 + 10x + 10 \), the first derivative is found by differentiating each term: \( g'(x) = 2x + 10 \). This tells us how the function's output changes as the input \( x \) changes.

To compute the first derivative:
  • Apply the power rule to \( x^2 \), which gives \( 2x \).
  • The derivative of \( 10x \) is \( 10 \).
  • Since the derivative of a constant is zero, \( \frac{d}{dx}(10) = 0 \).
  • Summarize to obtain \( g'(x) = 2x + 10 \).
Understanding the first derivative helps in finding the slope at any point on the curve, and is essential for locating critical points where the slope is zero.
Critical Points
Critical points of a function occur where its first derivative is zero or undefined. These points are potential locations of relative extrema (maximum or minimum). For \( g(x) = x^2 + 10x + 10 \), we first find its first derivative to be \( g'(x) = 2x + 10 \). Setting \( g'(x) = 0 \) to find the critical points provides:
  • \( 2x + 10 = 0 \)
  • Solve for \( x \): \( 2x = -10 \)
  • \( x = -5 \)
Hence, \( x = -5 \) is the critical point. This is the \( x \)-coordinate where the function's slope is zero. In the next steps, we analyze whether this point is a relative maximum or minimum.
Second Derivative Test
The second derivative test determines the concavity of the function and thus identifies whether a critical point is a relative maximum or minimum. For \( g(x) = x^2 + 10x + 10 \), we first found \( g'(x) = 2x + 10 \). Now, we find the second derivative \( g''(x) \):
  • Differentiate \( g'(x) = 2x + 10 \), which gives \( g''(x) = 2 \).
Since \( g''(x) = 2 \) is a positive constant (greater than 0), the function is concave upward at \( x = -5 \). This indicates that the critical point \( x = -5 \) is a relative minimum.

The second derivative test can be summarized as:
  • If \( g''(x) > 0 \), the function has a relative minimum at the critical point.
  • If \( g''(x) < 0 \), the function has a relative maximum at the critical point.
Relative Extrema
Relative extrema are the points where the function reaches a local maximum or minimum. Using the critical points and the second derivative test, we determine the nature of these extrema. For the given function \( g(x) = x^2 + 10x + 10 \) with the critical point \( x = -5 \):
  • We substitute \( x = -5 \) back into the original function to find the \( y \)-coordinate: \( g(-5) = (-5)^2 + 10(-5) + 10 \).
  • Simplify to get \( g(-5) = 25 - 50 + 10 = -15 \).
Thus, the relative minimum point is at \( (-5, -15) \). This point is where the function is lower than at any other nearby points. This completes the process of identifying and verifying the relative minimum using calculus methods.

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