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Chapter 2: Problem 37

In Problems \(37-44,\) find the linear approximation to the given functions at the specified points. Plot the function and its linear approximation over the indicated interval. $$ f(x)=x^{2} \text { at } a=2,[0,3] $$

Short Answer

Expert verified
The linear approximation at \( a = 2 \) is \( L(x) = 4x - 4 \).

Step by step solution

01

Find the function's derivative

First, we need to find the derivative of the function \( f(x) = x^2 \). The derivative represents the slope of the tangent line at any point \( x \), which is essential for the linear approximation. The derivative \( f'(x) \) is found by differentiating the function:\[ f'(x) = 2x \].
02

Evaluate the function and derivative at the point of tangency

Next, we need to evaluate the function and its derivative at the specified point \( a = 2 \). This will give us the y-coordinate and the slope at \( x = 2 \), which are needed to form the equation of the tangent line.Calculate \( f(2) \):\[ f(2) = 2^2 = 4 \].Calculate \( f'(2) \):\[ f'(2) = 2 \times 2 = 4 \].
03

Write the equation of the tangent line

Using the point-slope form of a line \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, \( x_1 = 2 \), and \( y_1 = 4 \), we form the equation of the tangent line.Substitute in the values:\[L(x) = 4 + 4(x - 2)\].Simplifying gives:\[L(x) = 4 + 4x - 8 = 4x - 4\].
04

Plot the function and its linear approximation

In this step, plot both the original function \( f(x) = x^2 \) and its linear approximation \( L(x) = 4x - 4 \) over the interval \([0, 3]\). The function will be a curve, while the linear approximation will be a straight line that touches \( f(x) \) at \( x = 2 \).This visual comparison illustrates how the linear approximation estimates the function near \( x = 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Line
In calculus, the tangent line is like a gentle hand that just touches the curve of a function at one point and shows the direction the curve is heading in at that spot. Imagine standing on a curve; the tangent line tells you the immediate path you'd follow if you kept walking without changing direction. So, what does a tangent line do in the world of calculus? Here鈥檚 how it helps in function approximation:
  • It provides a straight-line view or linear perspective of a curve's direction at a single point.
  • By assessing the tangent, we can understand and predict the behavior of a curve, especially in small neighboring intervals around a specific point.
The beauty of a tangent line is that it simplifies complex curves into straight lines, making them easier to calculate and understand in specific areas.
Derivative
The derivative of a function is an essential tool in calculus used to measure the instantaneous rate of change or the slope at any point on a curve. Much like checking how fast your car is moving at an exact moment, the derivative measures the speed of function changes.
  • For the function \( f(x) = x^2 \), the derivative \( f'(x) = 2x \) represents how quickly or slowly the curve of \( f(x) \) rises or falls.
  • The value of the derivative at a specific point gives the slope of the tangent line there.
  • Finding the derivative is a basic skill needed for solving more complex calculus problems.
Understanding derivatives allows us to closely examine the nature of changes within a function, leading us to more precise function approximation.
Function Approximation
Function approximation is a technique used to find a simple function that closely resembles the behavior of a more complex one. Linear approximation, a type of function approximation, utilizes the tangent line to estimate the value of a function near a particular point.When working with \( f(x) = x^2 \) at \( x = 2 \), we saw how the tangent line (\( L(x) = 4x - 4 \)) serves as a linear approximation:
  • The approximation is useful over a small range, giving a simpler equation for quick calculations.
  • It helps in making predictions without dealing with complicated functions.
  • It鈥檚 ideal for scenarios where exactness isn't necessary, and a reasonable estimate suffices.
Linear approximation is fundamental in calculus and is widely used in engineering, physics, and many other fields where function behavior near specific points is crucial.
Calculus Problem Solving
Calculus problem-solving often involves breaking down a complex system into manageable parts, much like solving a puzzle. The goal is to apply calculus tools to approach and solve real-world challenges step by step. Here's how you can tackle calculus problems efficiently:
  • Identify the core problem: Determine what you need to find and what information you have.
  • Use derivatives to understand rate changes and find slopes of tangent lines.
  • Apply linear approximation to simplify and make predictions about the function鈥檚 behavior near points of interest.
  • Visualize your results by plotting graphs to see the relationship between the function and its approximation.
Through such structured approaches, calculus not only becomes more accessible but also deeply rewarding as you arrive at solutions that describe the world with precision and simplicity.

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Most popular questions from this chapter

A business is prospering in such a way that its total (accumulated) profit after \(t\) years is \(1000 t^{2}\) dollars. (a) How much did the business make during the third year (between \(t=2\) and \(t=3) ?\) (b) What was its average rate of profit during the first half of the third year, between \(t=2\) and \(t=2.5 ?\) (The rate will be in dollars per year.) (c) What was its instantaneous rate of profit at \(t=2 ?\)

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). \(G(x)=\frac{2 x}{x^{2}-x}\)

The given limit is a derivative, but of what function and at what point? \(\lim _{x \rightarrow 3} \frac{x^{3}+x-30}{x-3}\)

Suppose that \(f(x+y)=f(x) f(y)\) for all \(x\) and \(y\). Show that if \(f^{\prime}(0)\) exists then \(f^{\prime}(a)\) exists and \(f^{\prime}(a)=f(a) f^{\prime}(0)\).

Use a CAS to do problems. Draw the graphs of \(f(x)=\cos x-\sin (x / 2)\) and its derivative \(f^{\prime}(x)\) on the interval [0,9] using the same axes. (a) Where on this interval is \(f^{\prime}(x)>0 ?\) (b) Where on this interval is \(f(x)\) increasing as \(x\) increases? (c) Make a conjecture. Experiment with other intervals and other functions to support this conjecture.
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