91Ó°ÊÓ

Derivatives are essential in determining how a function changes as its input changes. In this exercise, we are dealing with the function \(f(x) = \frac{6}{x^2}\). To find the derivative, \(f'(x)\), we use the power rule in conjunction with the rule for differentiating a quotient. The first step involves rewriting the function as \(6x^{-2}\). Using the power rule, where the derivative of \(x^n\) is \(nx^{n-1}\), we find that the derivative is \(f'(x) = -12x^{-3}\). Converting back, \(f'(x) = -\frac{12}{x^3}\).
You can always double-check your derivative by substituting back into the definition of a derivative, but remember these rules help simplify the process! Getting comfortable with derivatives forms the foundation for solving more complex calculus problems.

The tangent line touches a curve at precisely one point, and its slope at that point is equivalent to the derivative of the function. In this exercise, we calculate the slope of the tangent line by evaluating the derivative at the specific point \((2, \frac{3}{2})\).
Substituting \(x = 2\) into our derivative \(f'(x) = -\frac{12}{x^3}\), we find the slope \(f'(2) = -\frac{3}{2}\).
The general equation of a line is \(y = mx + c\), where \(m\) is the slope. For a tangent line at \(x = a\), our equation becomes \(y = f'(a)(x - a) + f(a)\). By inserting our values for \(m, a,\) and \(f(a)\), we find: \[T(x) = -\frac{3}{2}(x - 2) + \frac{3}{2}\].
This formula is essential for linear approximations.

Linear Approximation, also known as tangent line approximation, simplifies complex functions around points that are easier to calculate. We use the tangent line, which provides an estimate for the value of a function close to a known point. This is especially useful for functions that are difficult to compute directly.
To perform linear approximation using the tangent line equation \(T(x) = -\frac{3}{2}(x - 2) + \frac{3}{2}\), we substitute the values of \(x\) that are close to 2, like 1.9, 1.99, 2.01, and 2.1. By calculating \(T(x)\), you can estimate \(f(x)\) very close to our simple, calculated points. Linear approximation helps in understanding how a function behaves near known values.

Function substitution is a technique used throughout calculus and is fundamental when working with tangent lines and linear approximations. In this problem, it's part of applying our tangent line equation. We substitute different \(x\) values such as 1.9, 1.99, 2.01, and 2.1 into \(T(x) = - \frac{3}{2}(x - 2) + \frac{3}{2}\) to predict \(f(x)\).
This substitution helps us approximate \(f(x)\) without needing its exact calculation. It's particularly useful when the function is complex or computationally expensive to evaluate directly. Function substitution makes finding estimates quicker and helps build a better intuition about the function's behavior near certain points.