Question

Let \(x=1\) and \(\Delta x=0.01\). Find \(\Delta y\). \(f(x)=\frac{x}{x^{2}+1}\)

Short answer

The change in y(\(\Delta y\)) for the given function \(f(x)=\frac{x}{x^{2}+1}\) when \(x=1\) and \(\Delta x=0.01\) is approximately -0.004902.

Step-by-step solution

Substitute the given values into the function

First, calculate the values of the function \(f(x)\) and \(f(x+\Delta x)\). For \(f(x)\), substitute \(x=1\) into \(f(x)=\frac{x}{x^{2}+1}\), to get \(f(1)=\frac{1}{1^{2}+1}=0.5\). Now, for \(f(x+\Delta x)\), substitute \(x+\Delta x=1+0.01=1.01\) into the function, getting \(f(1.01)=\frac{1.01}{1.01^{2}+1}\). This is a decimal to be calculated in the next step.

Calculate Decimal Value

Calculate the numerical value of \(f(1.01)=\frac{1.01}{1.01^{2}+1}\) using a calculator to obtain the approximate value, which is about 0.495098.

Calculate Delta y

Now, to find \(\Delta y\), we can use the formula \(\Delta y = f(x+\Delta x) - f(x)\). Substituting the calculated values from previous steps, we will use \(\Delta y = f(1.01) - f(1)\), which simplifies to \(\Delta y = 0.495098 - 0.5\). The final calculation will result in the value of \(\Delta y\).

Key concepts

Calculus

Calculus is a branch of mathematics that deals with rates of change (differential calculus) and the accumulation of quantities (integral calculus). It forms the foundation for much of modern science and engineering. In the example provided, calculus is used to calculate the change in the y-value of a function, represented as \(\Delta y\), given a small change in the x-value, or \(\Delta x\). This is relevant to real-world applications such as calculating speed (a rate of change) or the area under a curve (an accumulation of quantities).

Understanding calculus allows students to solve problems in physics, economics, biology, and beyond, where dynamic changes and accumulative effects play a significant role.

Difference Quotients

Difference quotients are used in calculus to define the derivative of a function, which is a measure of how a function changes as its input changes. It's represented as \(\frac{f(x+\Delta x) - f(x)}{\Delta x}\) and simplifies to the derivative of the function when \(\Delta x\) approaches zero. The example exercise focuses on the change in the function's y-value, \(\Delta y\), but does not divide by \(\Delta x\).

Students can think of difference quotients as a way to approximate the slope of the curve at a point before they learn about derivatives. It's like zooming in on a curve at a specific point to see how steep it is.

Function Evaluation

When you evaluate a function, you're finding out what value it outputs for a given input. The process is simple: substitute the input value into the function's formula. In our example, we evaluated \(f(x)\) for \(x=1\) and \(f(x+\Delta x)\) for \(x=1\) and \(\Delta x=0.01\). By doing this, we obtained the respective values of the function which are required to calculate \(\Delta y\).

Evaluating functions correctly is essential in fields like computer science, where such evaluations are used in algorithms and complex computations. It's about plugging the right values in and doing the arithmetic to get a result.

Increment Method

The increment method, also known as the finite difference method, is a technique used to approximate derivatives and understand the behavior of functions. It involves calculating the increase or decrease in a function's value, \(\Delta y\), as a result of a small increment in the variable, \(\Delta x\). In the exercise, the increment method is applied by taking the function's output for \(x=1\) and comparing it to its output for a slightly increased \(x\) value of \(1.01\).

Using this method is fundamental in numerical analysis and applied mathematics, where it is often impossible or impractical to find exact changes. It allows for the practical modeling of complex systems and processes across different scientific disciplines.