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Find the linearization of \(f(x)=\sqrt{x+1}+\sin x\) at \(x=0 .\) How is it related to the individual linearizations of \(\sqrt{x+1}\) and \(\sin x\) at \(x=0 ?\)

Short Answer

Expert verified
The linearization of \(f(x)\) at \(x=0\) is \(L(x) = 1 + \frac{3}{2}x\), matching the sum of the individual linearizations of \(\sqrt{x+1}\) and \(\sin x\).

Step by step solution

01

Understanding the Linearization Concept

The linearization of a function at a point gives us the best linear approximation of the function near that point. Mathematically, the linearization of a function \(f(x)\) at \(x = a\) is given by:\[ L(x) = f(a) + f'(a)(x-a) \] We need to find the linearization of \(f(x) = \sqrt{x+1} + \sin x\) at \(x = 0\).
02

Find \(f(0)\)

Substitute \(x = 0\) into the function to find \(f(0)\):\[ f(0) = \sqrt{0+1} + \sin(0) = 1 + 0 = 1 \]
03

Calculate the Derivative \(f'(x)\)

Differentiate \(f(x) = \sqrt{x+1} + \sin x\):- The derivative of \(\sqrt{x+1}\) is \(\frac{1}{2\sqrt{x+1}}\).- The derivative of \(\sin x\) is \(\cos x\).Thus, \[ f'(x) = \frac{1}{2\sqrt{x+1}} + \cos x \]
04

Evaluate \(f'(0)\)

Substitute \(x = 0\) into the derivative to find \(f'(0)\):\[ f'(0) = \frac{1}{2\sqrt{0+1}} + \cos(0) = \frac{1}{2} + 1 = \frac{3}{2} \]
05

Formulate the Linearization \(L(x)\)

Using the linearization formula:\[ L(x) = f(0) + f'(0)(x - 0) = 1 + \frac{3}{2}x \] Thus, the linearization of \(f(x)\) at \(x=0\) is \(L(x) = 1 + \frac{3}{2}x\).
06

Compare with Individual Linearizations

Now, let's examine the linearizations of each individual function:- For \(\sqrt{x+1}\) at \(x=0\), \(L_1(x) = 1 + \frac{1}{2}x\).- For \(\sin x\) at \(x=0\), \(L_2(x) = x\).Adding these linearizations:\[ L_1(x) + L_2(x) = (1 + \frac{1}{2}x) + x = 1 + \frac{3}{2}x \]This matches the linearization of \(f(x)\), showing the linearization of a sum is the sum of the linearizations.

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Key Concepts

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

Understanding Derivatives
Derivatives are a fundamental concept in calculus, representing the rate at which a function changes. They provide us with a way to understand how a function's output alters based on changes in the input. When differentiating a function, we look for how small changes in parameters affect the function's value.
In our exercise, the derivative of the function \(f(x) = \sqrt{x+1} + \sin x\) was found to be \(f'(x) = \frac{1}{2\sqrt{x+1}} + \cos x\). This tells us how the function behaves locally, around any point \(x\).
For linearization at a specific point, say \(x=0\), we evaluate the derivative at this point to understand the slope of the tangent line. In this case, by substituting \(x = 0\), we get \(f'(0) = \frac{3}{2}\), indicating a steady increase in the function as \(x\) grows from zero.
Concept of Approximation
Approximation is a powerful mathematical tool that helps us simplify complex functions into more easily understandable forms. When dealing with functions like \(f(x) = \sqrt{x+1} + \sin x\), calculating or retrieving exact values can be cumbersome at times.
Linear approximation, or linearization, makes these functions more manageable by approximating them as a line at a specific point. It allows us to use the tangent line equation instead of the entire function for estimates near the point of interest.
  • Why Approximate? - To simplify complex calculations using simple linear expressions for small changes around a known value.
  • How it Works: - It ignores higher-order terms that have minimal impact on small intervals.
The beauty of linear approximation appears when we realize that linearizing complex sums like our function is equivalent to adding the linearizations of its parts, as shown in the exercise.
Exploring Function Analysis
Function analysis involves breaking down a function to understand its components, behavior, and characteristics. This includes recognizing how each simple function contributes to the behavior of the overall expression.
A function like \(f(x) = \sqrt{x+1} + \sin x\) can be seen as an aggregation of two primary expressions: \(g(x) = \sqrt{x+1}\) and \(h(x) = \sin x\). Each has its distinct behavior and properties. For example, \(g(x)\) involves a root function, which grows slower, while \(h(x)\) follows a periodic pattern due to its trigonometric nature.
  • Individual Analysis: Understanding how each part behaves helps predict the entire function's behavior.
  • Combination Insight: Linearizing each component individually and then summing gives insights on function behavior near the point.
By analyzing and then linearizing each separately, we see how combining these linearizations matches the original function's linear form. Thus, understanding specifics of each element helps us predict when combined.

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Most popular questions from this chapter

Oil spill An explosion at an oil rig located in gulf waters causes an elliptical oil slick to spread on the surface from the rig. The slick is a constant 9 in. thick. After several days, when the major axis of the slick is 2 mi long and the minor axis is 3\(/ 4\) mi wide, it is determined that its length is increasing at the rate of 30 \(\mathrm{ft} / \mathrm{hr}\) , and its width is increasing at the rate of 10 \(\mathrm{ft} / \mathrm{hr}\) . At what rate (in cubic feet per hour) is oil flowing from the site of the rig at that time?

Ships Two ships are steaming straight away from a point \(O\) along routes that make a \(120^{\circ}\) angle. Ship \(A\) moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship \(B\) moves at 21 knots. How fast are the ships moving apart when \(O A=5\) and \(O B=3\) nautical miles?

In Exercises \(35-40,\) write a differential formula that estimates the given change in volume or surface area. $$ \begin{array}{l}{\text { The change in the surface area } S=6 x^{2} \text { of a cube when the edge }} \\ {\text { lengths change from } x_{0} \text { to } x_{0}+d x}\end{array} $$

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2 $$

a. Find the tangent to the curve \(y=2 \tan (\pi x / 4)\) at \(x=1\) b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval \(\quad-2

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