Understanding the range of functions is a fundamental concept in calculus and algebra. In this article, we will explore the range of the function f(x) x2 - x1. We will use both algebraic and calculus methods to find the range of this function. This will not only help in understanding the behavior of the function but also provide insights into essential mathematical concepts.

Introduction

Given the function f(x) x2 - x1. The goal is to determine the range of this function. The range of a function is the set of all possible output values (y-values) of the function. To find the range of f(x), we will first analyze the expression for f(x) and then employ both algebraic and calculus methods to find the range.

Algebraic Approach

Let's first use the expression for the function to determine the range without resorting to calculus.

Given: [f(x) x^2 - 0.52075]

To find the range of this function, we need to consider the behavior of the quadratic function (x^2 - 0.52075). The term (x^2) is a parabola that opens upwards, and the value of -0.52075 shifts the parabola vertically.

Let's rewrite the function as:

[f(x) x^2 - 0.52075]

The smallest value of (x^2) is 0, which occurs when (x 0). Adding the constant -0.52075 to this value, we get:

[f(0) 0 - 0.52075 -0.52075]

Since the function (x^2) can take any non-negative value, adding -0.52075 to these values means the minimum value of the function is -0.52075. Therefore, the range of the function is:

[text{Range} [-0.52075, infty)]

Calculus Approach

To confirm our findings using calculus, let's analyze the function f(x) x2 - x1 using calculus techniques.

First, we find the first derivative of the function to find the critical points:

[f(x) x^2 - 0.52075]

[f'(x) 2x - 1]

Setting the first derivative to zero to find the critical points:

[2x - 1 0]

[x frac{1}{2}]

Next, we need to determine the second derivative to check the nature of the critical points:

[f''(x) 2]

Since the second derivative (f''(x) 2) is positive, the critical point (x frac{1}{2}) is a minimum.

Now, let's evaluate the function at the critical point:

[fleft(frac{1}{2}right) left(frac{1}{2}right)^2 - 0.52075]

[fleft(frac{1}{2}right) frac{1}{4} - 0.52075 -0.27075]

Since the function is a parabola that opens upwards, the minimum value of the function is -0.27075. Therefore, the range of the function is:

[text{Range} [-0.27075, infty)]

Conclusion

In conclusion, using both algebraic and calculus approaches, we have found the range of the function (f(x) x^2 - 0.52075) to be ([-0.27075, infty)). This detailed analysis provides a comprehensive understanding of the function's behavior and the range of possible output values.

Keywords: range of function, calculus, minimum value, critical points, algebra