Problem 44 - Calculus Early Transcendentals

Question

Let \(f(x)=x^{2}+p x+q .\) Find the values of \(p\) and \(q\) such that \(f(1)=3\) is an extreme value of \(f\) on \([0,2] .\) Is this value a maximum or minimum?

Step-by-Step Solution

Answer

For \(p = -2\) and \(q = 4\), \(f(1) = 3\) is a minimum.

Step 1: Understand the problem

We have a quadratic function \(f(x) = x^2 + px + q\) and we need to find \(p\) and \(q\) such that \(f(1) = 3\). This means when \(x = 1\), the function equals 3, making this an extreme value on the interval [0, 2]. We also need to determine if this value is a maximum or a minimum.

Step 2: Setting up the equation for the given condition

From the condition \(f(1) = 3\), we substitute into the function: \[1^2 + p(1) + q = 3\] which simplifies to: \[1 + p + q = 3\].Rearranging gives: \[p + q = 2\]. (Equation 1)

Step 3: Finding the condition for extremity

For \(f(x)\) to have an extreme value at \(x = 1\), the derivative \(f'(x)\) must be zero at this point. The derivative of \(f(x) = x^2 + px + q\) is: \[f'(x) = 2x + p\]. Setting \(f'(1) = 0\), we get: \[2(1) + p = 0\] which simplifies to: \[2 + p = 0\]. Solving for \(p\) gives: \[p = -2\]. (Equation 2)

Step 4: Solving for \(q\)

Using Equation 2, \(p = -2\), substitute in Equation 1: \[-2 + q = 2\] Solving for \(q\), we get: \[q = 4\].

Step 5: Determine maximum or minimum

The second derivative, \(f''(x)\), helps to determine if the extreme value is a maximum or minimum. Calculate \(f''(x) = 2\).Since \(f''(x) > 0\), this means \(f(x)\) is concave up at \(x = 1\), confirming that \(f(1) = 3\) is a minimum.

Key Concepts

Extreme Values

In mathematics, an extreme value of a function refers to either a maximum or a minimum point. For quadratic functions like \(f(x) = x^2 + px + q\), which are parabolic in nature, these points are often found at the vertex of the parabola.
An extreme value is crucial because it represents either the highest or lowest point on the graph over a specific interval. In our exercise, we look for extreme values on the interval \([0, 2]\).
To find an extreme value, we need the derivative of the function to be zero at that point. This is based on the idea that at high and low turning points of a continuous curve, the slope (or rate of change) is flat, which corresponds to the derivative being zero.
Extreme values can offer insightful information about the behavior of the function. They tell us where the graph changes direction, which can be useful for understanding the overall shape and pattern of a function.

Derivatives

Derivatives play a vital role in calculus and are used to measure how a function changes as its input changes. More specifically, a derivative gives the slope of a function at any given point.
For our quadratic function \(f(x) = x^2 + px + q\), we find its derivative to understand how the function behaves around the point \(x = 1\). The derivative is given by \(f'(x) = 2x + p\).
To locate extreme values, such as maxima or minima in the interval considered, we often set this first derivative to zero. This calculation let's us confirm points where the function stops increasing or decreasing. In our case, setting \(f'(1)=0\) leads to solving \(2 + p = 0\), finding \(p = -2\).
Derivatives not only pave the way to locate extreme points but also indicate intervals where the function is increasing or decreasing, giving a more complete picture of the function's behavior.

Concavity Test

The concavity test is an important tool to determine whether an extreme point is a maximum or a minimum.
The idea is simple: by examining the second derivative of a function, known as \(f''(x)\), we can determine the curvature, or concavity, around the extreme point. For quadratic functions like ours, the second derivative is constant; here, \(f''(x) = 2\).
If \(f''(x) > 0\), the graph is concave up, resembling a bowl shape. This indicates that the vertex, or extreme point, is a minimum. Conversely, if \(f''(x) < 0\), the graph is concave down, like an upside-down bowl, indicating a maximum.
In this exercise, since \(f''(x) = 2\) is positive, the extreme value at \(x = 1\) is indeed a minimum. This test gives a quick and definitive answer to whether we have a high or low turning point at an extreme value.