Finding Linearization L(x) For G(x) = Xf(x^2) At X=2
In the realm of calculus, linearization stands as a powerful tool for approximating the behavior of a function near a specific point. This technique is particularly useful when dealing with complex functions, as it allows us to replace them with simpler linear approximations, making calculations and analysis significantly easier. In this comprehensive guide, we will delve into the process of finding the linearization L(x) of the function g(x) = xf(x^2) at x = 2, given the following information:
- f(2) = 0
- f'(2) = 10
- f(4) = 5
- f'(4) = -2
This exploration will not only provide a step-by-step solution but also shed light on the underlying concepts and their practical applications. Understanding linearization is crucial for various fields, including physics, engineering, and economics, where approximating complex models with linear ones is a common practice. The problem at hand combines the concepts of derivatives, chain rule, and the definition of linearization, making it an excellent exercise for solidifying your calculus skills. We will break down each step, ensuring a clear understanding of the process and the reasoning behind it. This guide aims to equip you with the knowledge and confidence to tackle similar problems and appreciate the versatility of linearization in mathematical problem-solving.
Understanding Linearization
Before we dive into the specifics of this problem, let's first establish a solid understanding of what linearization entails. The linearization of a function g(x) at a point x = a is essentially the equation of the tangent line to the graph of g(x) at that point. This tangent line provides a linear approximation of the function's behavior in a small neighborhood around x = a. The formula for the linearization L(x) is given by:
L(x) = g(a) + g'(a)(x - a)
Here, g(a) represents the value of the function at x = a, and g'(a) denotes the derivative of the function evaluated at x = a. The derivative, g'(a), gives us the slope of the tangent line, while g(a) gives us a point on the line. The term (x - a) represents the change in x from the point of linearization. This formula is derived from the point-slope form of a line, which is a fundamental concept in coordinate geometry. The accuracy of the linear approximation depends on how close x is to a. The closer x is to a, the better the approximation. This is because the tangent line closely resembles the function's behavior in a small interval around the point of tangency. However, as x moves further away from a, the approximation may become less accurate, as the function's curvature becomes more significant. In practical applications, linearization is often used to simplify complex functions, making them easier to analyze and manipulate. For example, in physics, small-angle approximations are a form of linearization used to simplify trigonometric functions. In optimization problems, linearization can help in finding local optima by approximating the objective function with a linear one. Understanding the limitations of linearization is as important as understanding its applications. It is crucial to be aware of the interval in which the approximation is valid and to consider the potential error introduced by the approximation. The error typically increases as we move further away from the point of linearization. Therefore, it is essential to choose the point of linearization a carefully, depending on the specific problem and the desired level of accuracy.
Applying the Concept to Our Problem
In our specific problem, we are tasked with finding the linearization L(x) of the function g(x) = xf(x^2) at x = 2. This means we need to determine g(2) and g'(2), and then plug these values into the linearization formula. The given information includes values of the function f and its derivative f' at specific points, which we will use to compute the required values. The first step is to find g(2). Substituting x = 2 into the expression for g(x), we get:
g(2) = 2f(2^2) = 2f(4)
We are given that f(4) = 5, so:
g(2) = 2 * 5 = 10
Thus, we have found the value of the function g(x) at the point of linearization, x = 2. This value will be the y-intercept of our linear approximation, shifted appropriately by the g'(2)(x - 2) term. Next, we need to find the derivative of g(x), denoted as g'(x). This requires applying the product rule and the chain rule of differentiation. The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Applying these rules to g(x) = xf(x^2), we get:
g'(x) = (1)f(x^2) + x * f'(x^2) * (2x)
Simplifying this expression, we have:
g'(x) = f(x^2) + 2x2f'(x2)
Now, we need to evaluate g'(x) at x = 2 to find g'(2). Substituting x = 2 into the expression for g'(x), we get:
g'(2) = f(2^2) + 2(22)f'(22) = f(4) + 8f'(4)
We are given that f(4) = 5 and f'(4) = -2, so:
g'(2) = 5 + 8(-2) = 5 - 16 = -11
We have now found the derivative of g(x) evaluated at the point of linearization, x = 2. This value represents the slope of the tangent line to the graph of g(x) at x = 2, and it is a crucial component of the linear approximation. With both g(2) and g'(2) determined, we are ready to construct the linearization L(x).
Calculating the Derivative g'(x)
As we established in the previous section, finding the derivative g'(x) is a critical step in determining the linearization of g(x). Given that g(x) = xf(x^2), we must employ both the product rule and the chain rule to correctly differentiate this function. The product rule is essential because g(x) is a product of two functions, x and f(x^2). The chain rule is necessary because f is a composite function, with the outer function being f and the inner function being x^2. Let's break down the process step by step to ensure clarity and accuracy. First, recall the product rule:
(uv)' = u'v + uv'
where u and v are functions of x, and the prime notation (') denotes differentiation with respect to x. In our case, let u(x) = x and v(x) = f(x^2). Then, u'(x) = 1. To find v'(x), we need to apply the chain rule. The chain rule states:
[f(g(x))]' = f'(g(x)) * g'(x)
Applying the chain rule to v(x) = f(x^2), we get:
v'(x) = f'(x^2) * (2x)
Now we have all the components to apply the product rule to find g'(x):
g'(x) = u'(x)v(x) + u(x)v'(x)
Substituting our expressions for u'(x), v(x), u(x), and v'(x), we get:
g'(x) = (1)f(x^2) + x * f'(x^2) * (2x)
Simplifying this expression, we arrive at:
g'(x) = f(x^2) + 2x2f'(x2)
This is the derivative of g(x). It is essential to understand the application of both the product rule and the chain rule in this context. Each rule plays a vital role in correctly differentiating the composite function. The product rule handles the product of x and f(x^2), while the chain rule handles the differentiation of the composite function f(x^2). With g'(x) in hand, we can now proceed to evaluate it at x = 2, which is the next step in finding the linearization.
Constructing the Linearization L(x)
Having computed g(2) = 10 and g'(2) = -11, we are now fully equipped to construct the linearization L(x) of the function g(x) = xf(x^2) at x = 2. Recall the formula for linearization:
L(x) = g(a) + g'(a)(x - a)
In our case, a = 2, so the formula becomes:
L(x) = g(2) + g'(2)(x - 2)
Substituting the values we found for g(2) and g'(2), we get:
L(x) = 10 + (-11)(x - 2)
This expression represents the equation of the tangent line to the graph of g(x) at x = 2. It provides a linear approximation of the function's behavior in the vicinity of x = 2. To simplify the expression, we can distribute the -11 and combine like terms:
L(x) = 10 - 11x + 22
Combining the constant terms, we get:
L(x) = -11x + 32
This is the final form of the linearization L(x). It is a linear function, as expected, and it provides a good approximation of g(x) for values of x close to 2. The slope of this line is -11, which is the value of the derivative g'(2), and the y-intercept is 32. The linearization L(x) can be used to estimate the value of g(x) for x values near 2. For instance, if we wanted to approximate g(2.1), we could use L(2.1) instead:
L(2.1) = -11(2.1) + 32 = -23.1 + 32 = 8.9
This approximation gives us a quick and easy way to estimate the value of g(2.1) without having to directly evaluate the more complex function g(x). The accuracy of this approximation will depend on how close 2.1 is to 2. In general, the closer the x value is to the point of linearization, the more accurate the approximation will be. However, it is essential to keep in mind that linearization is an approximation, and there will be some error associated with it. The error typically increases as we move further away from the point of linearization. Therefore, it is crucial to use linearization judiciously and to be aware of its limitations. In summary, we have successfully constructed the linearization L(x) = -11x + 32 of the function g(x) = xf(x^2) at x = 2. This linear function provides a valuable approximation of g(x) near x = 2 and can be used to simplify calculations and analysis.
Final Answer: L(x) = -11x + 32
In conclusion, we have successfully found the linearization L(x) of the function g(x) = xf(x^2) at x = 2, given the provided information about f(x) and its derivative. By systematically applying the concepts of derivatives, the product rule, the chain rule, and the definition of linearization, we arrived at the solution:
L(x) = -11x + 32
This linear function represents the tangent line to the graph of g(x) at x = 2 and provides a useful approximation of the function's behavior in the vicinity of this point. The process involved several key steps. First, we established the fundamental concept of linearization and its importance in approximating complex functions. Then, we identified the specific requirements of the problem, namely finding g(2) and g'(2). We computed g(2) by directly substituting x = 2 into the expression for g(x) and using the given value of f(4). The next crucial step was to find the derivative g'(x), which required a careful application of the product rule and the chain rule. We broke down this process step by step to ensure clarity and accuracy. Once we had the expression for g'(x), we evaluated it at x = 2 using the given values of f(4) and f'(4). Finally, with both g(2) and g'(2) in hand, we plugged these values into the linearization formula and simplified the expression to obtain the final answer. This problem serves as a valuable example of how linearization can be used to approximate complex functions with simpler linear ones. The ability to find linearizations is a fundamental skill in calculus and has numerous applications in various fields, including physics, engineering, and economics. Understanding the underlying concepts and mastering the techniques involved is essential for success in these areas. We hope this comprehensive guide has provided a clear and thorough understanding of the process of finding the linearization of a function and has equipped you with the knowledge and confidence to tackle similar problems in the future. The final answer, L(x) = -11x + 32, encapsulates the essence of linear approximation and highlights the power of calculus in simplifying complex mathematical problems.
