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Chapter 2: Problem 2

Find the relative extrema of each function, if they exist. List each extremum along with the \(x\) -value at which it occurs. Then sketch a graph of the function. $$f(x)=x^{2}+6 x-3$$

Short Answer

Expert verified
The relative minimum occurs at \( x = -3 \) with a value of \(-12\).

Step by step solution

01

Find the derivative

The first step is to find the first derivative of the function to determine the critical points. Given the function \( f(x) = x^2 + 6x - 3 \), the derivative is calculated as: \[ f'(x) = 2x + 6 \]
02

Set the derivative equal to zero

To find the critical points, set the first derivative equal to zero and solve for \( x \): \[ 2x + 6 = 0 \] Solving this equation gives: \[ x = -3 \]
03

Determine if the critical point is a relative extremum

To identify whether the critical point is a relative minimum or maximum, use the second derivative test. Compute the second derivative of the function: \[ f''(x) = 2 \]. Since \(f''(x) > 0\) for all \(x\), the function is concave up, indicating that \(x = -3\) is a relative minimum.
04

Find the function value at the critical point

Substitute \( x = -3 \) back into the original function to find the \( y \)-value: \[ f(-3) = (-3)^2 + 6(-3) - 3 \], which simplifies to \[ f(-3) = 9 - 18 - 3 = -12 \].
05

State the relative extremum

The function \( f(x) \) has a relative minimum at \( x = -3 \) with a function value of \( -12 \). So, the relative minimum point is \((-3, -12)\).
06

Sketch the graph

The function \( f(x) = x^2 + 6x - 3 \) is a parabola opening upwards. Plot the vertex at the point \((-3, -12)\), and draw the parabola opening upwards to sketch the graph.

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Key Concepts

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

derivative
In calculus, a derivative measures how a function changes as its input changes. It provides the slope of the function at any given point.
To find the critical points of a function, we must first compute its derivative. For our example function, which is \( f(x) = x^2 + 6x - 3 \), the derivative can be found as follows:
\[ f'(x) = 2x + 6 \]
This derivative tells us how steep the function is at any point. The next step is setting this derivative equal to zero to find critical points.
second derivative test
The second derivative test helps us determine the nature of critical points, i.e., whether they are relative minima, maxima, or saddle points.
Once we have identified a critical point by setting the first derivative to zero, as shown:

\[ 2x + 6 = 0 \rightarrow x = -3 \]

we need the second derivative to identify the concavity of the function at this point. For our function, the second derivative is:
\[ f''(x) = 2 \]
Since \( f''(x) = 2 \) is always positive, the function is concave up. Thus, the critical point at \( x = -3 \) is a relative minimum.
critical points
Critical points are where the first derivative of a function is zero or undefined.
These points are candidates for identifying local extrema (peaks or troughs).
For the function \( f(x) = x^2 + 6x - 3 \), we found the critical point by solving \[ 2x + 6 = 0, \text{ which yields } x = -3 \]
To verify if this critical point is a relative extremum, we would then use the second derivative test.
This point could be crucial in determining the curve's shape and for plotting the function accurately on a graph.
concavity
Concavity refers to the direction a function curves. If a function is concave up, it resembles a 'U' shape, and if concave down, it resembles an 'n' shape.
The second derivative tells us about the concavity. If the second derivative is positive at a point, the function is concave up; if negative, it is concave down.
For our quadratic function \( f(x) = x^2 + 6x - 3 \), the second derivative is constant at:
\[ f''(x) = 2 \]
This positive value indicates that the function is concave up across its entire domain, helping us determine the nature of points along its curve.
Specifically, the critical point found at \( x = -3 \) is a relative minimum.

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