Rivalry2

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Original 2026-01-01

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1 If f(-x)=f(x) for every value of x in the domain of f, then the function is said to be even; if \(f(-x)=-f(x)\) for every value of x in the domain of f, then the function is said to be<a>odd</a>. For all values of x in the domain of f, the zero function is the only function that satisfies both of these requirements simultaneously. The zero function is both odd and even. Since the output is always the same,

2 \(\ f(-x)=f(x)=-f(x)=0 \quad\text{if } f(x)=0 \ \) It is the zero function.

3 How to Find Zeros of FunctionWhen real, complex, or imaginary values are entered into a function, the function is equal to zero. Depending on the type of function, there are different ways to find its zeros. When describing a function or creating its graph, zeros are important.

4 Finding a function’s zeros can be done in a number of ways, including:

5 <ul><li>Graphical Method</li>

6 <li>Factorization Method</li>

7 <li>Quadratic Formula Method</li>

8 <li>Newton-Raphson Method</li>

9 <li>Bisection Method</li>

10 <li>Graphical Method</li>

11 </ul>A function’s graph can be used to determine its zeros. We are aware that the value of f(x) is zero for any zero of the function’s root. Therefore, the zeros of the function are represented by the x coordinate of each point where the graph of the function crosses the x-axis. For example, the graph of the function \(\ f(x) = x^2 - \frac{25}{16} \ \) intersects the x-axis. The zeros of the given function are \(x = \frac{5}{4}\) and \(x =\frac{ -5}{4}\), where the function equals zero.

12 Factorization methodWhen a function can be broken down into smaller parts, the factorization method is a helpful technique. To use the factorization method and to find the function’s zeros, just break the function down into simple<a>factors</a>,<a>set</a>each factor to zero, and solve them. Equating both factors to zero is based on the step that one or both factors must be zero when the<a>product</a>of the<a>expressions</a>is zero. For example, the factorization method can be used to find the zeros of the function\( f(x)=x^2-6x+5\) as follows:

13 \(f(x)=x^2-6x+5\) \(f(x)= (x-5)(x-1)\) put, \(x-5=0\) and \(x-1=0\) So, x=5 or x=1

14 Quadratic formula methodThe quadratic formula is one of the best algebraic techniques for determining a function’s roots. The roots of a quadratic function can be found with this method. Simply enter the values of a, b, and c in the<a>quadratic</a>formula, where a, b, and c stand for the<a>coefficients</a>of \(x^2\), x, and the constant term, respectively, to determine the roots of a quadratic function using this method.

15 \(\ \text{Root} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ \)

16 Or

17 \(\ \text{Root} = \frac{-b - b^2 - 4ac}{2a} \ \)

18 Using this technique, one can find the real roots and the imaginary roots:

19 \(\ \text{Real roots exist if } b^2 - 4ac \ge 0 \ \), real roots

20 \(\ \text{Imaginary roots exist if } b^2 - 4ac < 0 \ \)

21 Newton-Raphson methodOne of the most popular methods for finding the roots of a real-valued function is the Newton-Raphson method. It bears the names of Joseph Raphson and Sir Isaac Newton. The Newton-Raphson method approximates the subsequent iteration (\(x_1\)), which is near the root, after assuming the first iteration (\(x_0\)). The Newton-Raphson method works as follows:

22 Given that \(x_0\) is the approximate root of the<a>equation</a>, let \(\ x_1 = x_0 + h \ \) be the root of the function.

23 Then, \(\ f(x_1) = 0 \text{ or } f(x_0 + h) = 0 \ \),

24 Applying Taylor’s theorem, expand the equation above.

25 \(\ f(x_0) + h f'_1(x_0) + \ldots = 0 \ \)

26 Thus,

27 \(\ h = -\frac{f(x_0)}{f'(x_0)} \ \)or we could say

28 \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \)

29 Similarly, we can determine additional approximations for \(x_2\), \(x_3\), \(x_4\), etc.

30 Or

31 \(X_{n+1} = X_n - \frac{f(X_n)}{f'(X_n)} \)

32 This is called the Newton-Raphson formula.

33 Bisection method The root of a<a>polynomial</a>function within a specified interval can be found using the bisection method. Until the right answer is found, this method works by dividing the interval into smaller sub-intervals. For the functions that are continuous within a specified interval, this method is correct. Because of its strong similarity to binary search, it is also referred to as the interval having method or binary search method. The bisection method operates as follows:

34 Let x and y be such that x<y and f(x) x f(y)<0. Find the midpoint of x and y in each interval (let’s say it’s m). If m is the right root, then f(m) = 0. Thus, we obtain the necessary root m.

35 If not, the interval will be split into two sections: x to m and m to y. We will now select the interval based on the value of f(m) as follows:

36 Select the interval from m to y if f(m) < 0. Because x < m < y. Select the interval from x to m if f(m) > 0. Because x < m < y. Continue repeating steps one through three until the right answer is found.