Problem 71 - Single Variable Calculus

Question

Let $$ f(x)=\left\\{\begin{array}{ll} -x+2 & \text { if }-2 \leq x<0 \\ -\left(x^{2}+2\right) & \text { if } 0 \leq x \leq 2 \end{array}\right. $$ Does \(f\) have a zero in the interval \([-2,2] ?\) Explain your answer.

Step-by-Step Solution

Answer

No, the function \(f(x)\) does not have a zero in the interval \([-2,2]\). When analyzing both intervals for the respective pieces of \(f(x)\), there is no zero-point, and the function is discontinuous at x = 0, where the two pieces meet.

Step 1: Analyze the first piece of the function

We have: f(x) = -x + 2 if -2 ≤ x < 0 We can solve for x when f(x) = 0: 0 = -x + 2 x = 2 However, x = 2 is not within the given interval for this piece (-2 ≤ x < 0). Therefore, there is no zero-point in this interval.

Step 2: Analyze the second piece of the function

We have: f(x) = -(x^2 + 2) if 0 ≤ x ≤ 2 We can solve for x when f(x) = 0: 0 = -(x^2 + 2) x^2 + 2 = 0 x^2 = -2 Since x^2 must be non-negative, there is no real solution for x in this case. Thus, there is no zero-point in the interval [0, 2].

Step 3: Check the function's continuity at x = 0

Since we have analyzed both intervals, we are now left to check if the function is continuous at x = 0, the point of intersection. First, find the limits at 0 from the left and the right. The limit from the left (x < 0) is given by: \(\lim_{x\to0^-}(-x + 2)\) The limit from the right (x ≥ 0) is given by: \(\lim_{x\to0^+}(-(x^2 + 2))\) Now calculate the limits: \(\lim_{x\to0^-}(-x + 2) = -0 + 2 = 2\) \(\lim_{x\to0^+}(-(x^2 + 2)) = -(0^2 + 2) = -2\) Since the limits are different, the function is discontinuous at x = 0.

Step 4: Conclusion

The function f(x) does not have a zero-point in the interval [-2, 2]. We found that neither of the intervals for the respective pieces have a zero-point, and the function is discontinuous at x = 0, where the two pieces meet. Therefore, there is no zero-point for f(x) in the given interval [-2, 2].

Key Concepts

Zero of a Function

In mathematics, the "zero of a function" refers to any input value that results in the output of the function being zero. For a function \(f(x)\), the zero is a solution to the equation \(f(x) = 0\). Finding the zero of a function is important because it tells us where the graph of the function intersects the x-axis.
In the given piecewise function, we have two different expressions based on the interval of \(x\):

• -x + 2, when -2 ≤ x < 0
• -(x^2 + 2), when 0 ≤ x ≤ 2

The task is to determine if \(f(x) = 0\) for any \(x\) in the interval [-2,2]. However, as seen in the solution, there are no values of \(x\) such that \(f(x) = 0\) for either expression within their respective intervals. As a result, the function has no zero in the given interval.

Continuity

Continuity of a function at a point means that the function is defined at that point, the limit exists at that point, and the limit equals the function's value at that point. If a function is not continuous at a point, it means there is some kind of "break" or "gap." In this exercise, we check the continuity of the piecewise function at the point where the definitions change, which is \(x = 0\).
To assess continuity at \(x = 0\), we look at the limits of the function as \(x\) approaches 0 from both sides:

• The left-hand limit: \( \lim_{x\to0^-}(-x + 2) = 2\).
• The right-hand limit: \( \lim_{x\to0^+}(-(x^2 + 2)) = -2\).

Since these two limits are not equal, the function is discontinuous at \(x = 0\). This discontinuity means we cannot apply the intermediate value theorem around this point.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is an important concept in calculus. It states that if a function is continuous over a closed interval \([a, b]\) and \(f(a)e f(b)\), then the function takes every value between \(f(a)\) and \(f(b)\) at some point in the interval. This theorem is often used to prove that a function has a zero within a specific interval. However, in the given problem, the function \(f(x)\) is discontinuous at \(x = 0\). Because the IVT requires the function to be continuous over the entire interval \([-2, 2]\), we cannot use it to make any inferences in this case. This discontinuity breaks the required conditions for applying the Intermediate Value Theorem.

Function Analysis

Function analysis involves investigating the properties and behavior of a function. In this exercise, such analysis involves understanding the specific nature of the piecewise function and its components.
For the piecewise function

• The first piece, \(-x + 2\), is linear and decreasing for its interval \(-2 ≤ x < 0\). It has no zero within its designated interval.
• The second piece, \(-(x^2 + 2)\), is quadratic and always negative because the quadratic expression \(x^2 + 2\) is always positive for real numbers. Hence, \(-(x^2 + 2)\) never equals zero for \(0 ≤ x ≤ 2\).

Additionally, analyzing the continuity at the intersection point \(x = 0\) reveals a discontinuity. Collectively, this information confirms that \(f(x)\) has no zeros within the specified range. By conducting a thorough analysis, we gain deeper insights into the behavior and properties of the function.