Problem 67 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(\begin{aligned} f(x) &=x^{2}+1, &-2 \leq x<0 \\ &=-x^{2}-1, & 0 & \leq x \leq 2 . \end{aligned}\) Is there a point in the interval \([-2,2]\) at which \(f(x)=0 ?\)

Step-by-Step Solution

Answer

In the given piecewise function, \(f(x) = x^2 + 1\) for \(x \in [-2, 0)\) and \(f(x) = -x^2 - 1\) for \(x \in [0, 2]\). For both pieces, no value of x can yield \(f(x) = 0\) within the specified intervals, as the first piece has a minimum of 1 and the second piece has a maximum of -1. Therefore, there is no point in the interval \([-2, 2]\) at which the function f(x) equals 0.

Step 1: Identify the function within the given interval

The function is defined as: \(f(x) = \begin{cases} x^2 + 1 & -2 \leq x < 0 \\ -x^2 - 1 & 0 \leq x \leq 2 \end{cases}\) In the given interval of \([-2, 2]\) , the first piece \((x^2 + 1)\) corresponds to \(x \in [-2, 0)\) and the second piece \((-x^2 - 1)\) corresponds to \(x \in [0, 2]\). ---

Step 2: Solve the first piece for \(x \in [-2, 0)\)

For the first piece of the function, \(f(x) = x^2 + 1\), we want to find any \(x\) values within the interval \([-2, 0)\) such that \(f(x) = 0\). That is, we want to solve for: \[x^2 + 1 = 0\] We can see that since the \(x^2\) term is non-negative, the smallest this function can be is when \(x = 0\). Since this would equal to 1, no value of x will yield \(f(x) = 0\) for this interval. ---

Step 3: Solve the second piece for \(x \in [0, 2]\)

For the second piece of the function, \(f(x) = -x^2 - 1\), we want to find any \(x\) values within the interval \([0, 2]\) such that \(f(x) = 0\). That is, we want to solve for: \[-x^2 - 1 = 0\] Similar to the first piece, we can see that since the \(-x^2\) term is non-positive, the largest this function can be is when \(x = 0\). Since this would equal to -1, no value of x will yield \(f(x) = 0\) for this interval. ---

Step 4: Conclusion

Since we have shown that there is no solution for either piece of the function that would yield \(f(x) = 0\), we can conclude that there is no point in the interval \([-2, 2]\) at which the function f(x) equals 0.

Key Concepts

Finding Zeroes in Functions

To find the zeroes of a function means to determine the values of the variable (usually \(x\)) where the function's output equals zero. These points are significant because they represent where the graph of the function intersects the x-axis. In mathematical terms, you set the function equal to zero and solve for \(x\).

• For example, consider the function \(f(x) = x^2 + 1\). To find its zeroes, set it to zero, resulting in \(x^2 + 1 = 0\).
• However, solving this equation yields no real solution since \(x^2 + 1\) is always greater than zero for any real \(x\).
• This is similarly the case for \(f(x) = -x^2 - 1\), which is always less than zero, meaning no real zeroes exist because it never touches the x-axis.

Finding zeroes is crucial in analyzing the behavior of functions and can tell us much about the function's properties, such as symmetry, intervals of increase or decrease, or even maximum and minimum values.

Function Intervals

Function intervals tell us where specific behaviors happen within a function, such as where it increases, decreases, or remains constant. For piecewise functions like the one given, intervals dictate which part of the function to use.

• In our given function, \(f(x)\) is divided into two intervals: \([-2, 0)\) and \([0, 2]\).
• These intervals direct us on which equation to use for certain values of \(x\).
• The interval \([-2, 0)\) uses the expression \(x^2 + 1\), while the interval \([0, 2]\) employs \(-x^2 - 1\).

Understanding these intervals is key to analyzing the function's graph and ensures that computations are correctly categorized. Identifying function intervals simplifies complex problems by breaking them into smaller, easier-to-manage pieces.

Quadratic Functions

Quadratic functions are mathematical expressions of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). These functions represent parabolas when graphed, and their orientation depends on whether \(a\) is positive or negative.

• In this task, we examined both \(x^2 + 1\) and \(-x^2 - 1\), examples of quadratics.
• Each expression defines a parabola, albeit with different openings: \(x^2 + 1\) opens upwards while \(-x^2 - 1\) opens downwards.
• Despite being of quadratic form, neither expression has zeroes in its specified interval.

Quadratic functions are pivotal in many math applications. They are used to describe various phenomena in physics, engineering, and economics, often modeling critical points like peaks and troughs, which characterize many real-world processes.