Question

A function \(f\) is continuous on \([-2,2],\) and some of the values of \(f\) are shown below: $$\begin{array}{|c|c|c|c|}\hline x & -2 & 0 & 2 \\\\\hline f(x) & 3 & b & 4 \\\\\hline\end{array}$$ If \(f\) has only one root, \(r,\) on the closed interval \([-2,2],\) and \(r \neq 0,\) then a possible value of \(b\) is (A) -2 (B) -1 (C) 0 (D) 1

Short answer

Question: Based on the given information about the function \(f\), which is continuous on the closed interval \([-2,2]\), has only one root, and the root is not equal to \(0\), find a possible value for \(b\), the function value at \(0\). Answer: A possible value for \(b\) is \(-2\).

Step-by-step solution

Understand the Intermediate Value Theorem (IVT)

The Intermediate Value Theorem (IVT) states that if a function is continuous on a closed interval \([a,b]\), and if \(y_0\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in the interval \((a,b)\) such that \(f(c) = y_0\).

Apply IVT to the given function

Since \(f\) has only one root \(r\), and we know $rar, \end{cases}$$

Find possible values of \(b\)

We know that \(f(r) = 0\). So, we can apply the second and third equations from Step 4, which are derived using IVT. Case 1: If \(f(2) > 0\), then: $$f(0) = b < f(r) = 0$$ In this case, \(b\) is negative. Since \(f(-2)<0\), the only possible value from the options is (A) -2. Case 2: If \(f(-2) < 0\), then: $$f(0) = b > f(r) = 0$$ In this case, \(b\) is positive. Since \(f(2)>0\), the only possible value from the options is (D) 1. Both \(b=-2\) and \(b=1\) are possible values for \(b\). However, the question asks for only one possible value of \(b\). So, we can choose either one of them. For this solution, we choose: $$\boxed{\text{(A) } -2}$$

Key concepts

Continuous Function

When a function is described as continuous on an interval, it means that the graph of the function is unbroken or smooth over that interval. This characteristic ensures that there are no gaps, jumps, or holes as you trace the graph with your pen.
For example, consider the function values given in the problem:

• At \(x = -2\), \(f(x) = 3\)
• At \(x = 0\), \(f(x) = b\)
• At \(x = 2\), \(f(x) = 4\)

All these values lie on a continuous curve. This property is crucial for applying the Intermediate Value Theorem (IVT), which will be discussed later.
The continuity ensures that the function does not skip any value between \( f(-2) \) and \( f(2) \), meaning every y-value between 3 and 4 is hit by some x-value from the interval \([-2,2]\).

Closed Interval

A closed interval, denoted as \([-2, 2]\), includes all the numbers from -2 to 2, including the endpoints themselves. This is different from an open interval, which would exclude the endpoints.
Closed intervals are important in mathematical concepts like continuity, as they often set the boundaries for examining the behavior of functions.
In the exercise, the function is examined over the closed interval \([-2, 2]\), confirming that the function is continuous over this entire range. This inclusion of endpoints ensures that the function's behavior is fully observed from start to finish, and is essential for the application of the IVT, which relies on function values across a fully inclusive interval.

Root of a Function

The root of a function, also known as a zero, is a value of x for which \(f(x) = 0\). Finding the root is like asking, "Where does this function hit the x-axis?"
In our exercise, it is noted that the function \(f\) has only one root, \(r\), within the interval \([-2,2]\), and importantly, \(r eq 0\). This condition tells us that the graph of the function intersects the x-axis exactly once within this range.
Using the Intermediate Value Theorem, we can deduce that there must be a point where the function changes from positive to negative, or vice versa, crossing zero. Knowing the function’s endpoints and the value \(b\) helps in confirming this root.

Function Values

Function values are essentially the y-values that the function assumes for specific x-values, giving insights into the graph's shape and position. In the provided exercise, you have the following values:

• \(f(-2) = 3\)
• \(f(0) = b\)
• \(f(2) = 4\)

These values play a crucial role in applying the Intermediate Value Theorem, as they define the y-range of the function over the closed interval \([-2,2]\).
If the problem states there exists a root \(r\) somewhere in this range, then the function changes sign between these points. \(b\) stays a mystery value that helps determine either a positive or negative crossing.
An understanding of these values is essential in pinpointing how and why the root exists within the interval.