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Chapter 24: Problem 39

Solve the given linearization problems. Show that the linearization of \(f(x)=(1+x)^{k}\) at \(x=0\) is \(L(x)=1+k x\)

Short Answer

Expert verified
The linearization of \( f(x) = (1+x)^k \) at \( x = 0 \) is \( L(x) = 1 + kx \).

Step by step solution

01

Understand Linearization

Linearization is the approximation of a function near a point using the tangent line at that point. The linearization of a function \( f(x) \) around a point \( a \) is given by the formula \( L(x) = f(a) + f'(a)(x-a) \). Here, \( a = 0 \).
02

Find the Value of the Function at \( x = 0 \)

Evaluate the function \( f(x) = (1+x)^k \) at \( x = 0 \).\[ f(0) = (1+0)^k = 1^k = 1 \]
03

Compute the Derivative of the Function

Find the derivative of \( f(x) = (1+x)^k \) using the power rule.\[ f'(x) = k(1+x)^{k-1} \]
04

Evaluate the Derivative at \( x = 0 \)

Now, substitute \( x = 0 \) in the derivative to find \( f'(0) \).\[ f'(0) = k(1+0)^{k-1} = k \]
05

Write the Linearization Formula

Substitute the values from Step 2 and Step 4 into the linearization formula:\[ L(x) = f(0) + f'(0)(x - 0) = 1 + kx \]
06

Conclude the Calculation

The linearization \( L(x) = 1 + kx \) is the approximation of \( (1+x)^k \) near \( x = 0 \). This shows how the tangent line provides a simplified linear expression for the function near this point.

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Key Concepts

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

Derivative Calculation
Calculating derivatives is a crucial step in understanding how functions behave. The derivative of a function at a specific point can tell us about the rate of change at that point. In this case, we are dealing with the exponential function \(f(x) = (1+x)^k\). To find the derivative of this function, it involves basic differentiation rules.
  • First, understand that the derivative tells us how the function changes as \(x\) changes. You can think of it as the slope of the function at any given point.
  • For our function, the derivative is calculated with respect to \(x\) using the Power Rule (which we will talk about in more detail in the next section).
The derivative of \((1+x)^k\) is calculated to be \(f'(x) = k(1+x)^{k-1}\). This derivative shows how the function changes as \(x\) varies near a particular point, and it plays a crucial role when we move on to find the tangent line or the linearization.
Power Rule
The Power Rule is a basic differentiation rule that simplifies the process of finding derivatives of polynomial functions. The rule states that if you have a function of the form \( f(x) = x^n \), then its derivative is \( f'(x) = nx^{n-1} \).
  • Apply the rule: For the function \((1+x)^k\), consider it similar to a simple polynomial.
  • The derivative of \( (1+x)^k \) then becomes \( k(1+x)^{k-1} \).
This result is achieved by applying the Power Rule directly. It essentially reduces the power of the term by one and multiplies it by the original power, \(k\), which gives us \(k(1+x)^{k-1}\). This makes differentiation almost effortless when dealing with polynomial-like functions.
Linear Approximation
Linear approximation, also known as linearization, is a technique used to approximate a complex function with a simple linear function. Imagine trying to approximate a curvy road with a straight path.
  • We achieve this by using the tangent line, which closely matches the function at the point of interest.
  • For the function \((1+x)^k\), the linear approximation is derived using the tangent line at a specific point, here \(x = 0\).
Linearization gives us \( L(x) = f(0) + f'(0)(x-0) \). By substituting \( f(0) = 1 \) and \( f'(0) = k \), we get \( L(x) = 1 + kx \). This linear function approximates \((1+x)^k\) near \(x = 0\), providing us with a simpler expression that is easier to work with. This approximation can be used for quick estimates when exact calculations are not necessary.
Tangent Line at a Point
The tangent line is a straight line that touches a curve at a single point. It represents the instantaneous rate of change of the function at that particular point.
  • To find this line, you need both the value of the function and its derivative at the point.
  • For the function \((1+x)^k\) at \(x = 0\), the value \(f(0)\) is 1, and the slope of the tangent line, \(f'(0)\), is \(k\).
The equation for the tangent line then becomes \( y = 1 + kx \), which is the linearization formula. This line gives us a direct way to see how the function is changing exactly at \(x = 0\), and it simplifies our understanding of the function over small intervals.

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Most popular questions from this chapter

In Exercises \(43-54,\) sketch the indicated curves by the methods of this section. You may check the graphs by using a calculator. A computer analysis shows that the thrust \(T\) (in \(\mathrm{kN}\) ) of an experimental rocket motor is \(T=20+9 t-4 t^{3},\) where \(t\) is the time (in min) after the motor is activated. Sketch the graph for the first \(2 \mathrm{min}\)

In Exercises \(43-54,\) sketch the indicated curves by the methods of this section. You may check the graphs by using a calculator. A rectangular planter with a square end is to be made from \(64 \mathrm{ft}^{2}\) of redwood. Express the volume of soil the planter can hold as a function of the side of the square of the end. Sketch the curve of the resulting function.

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A company finds that there is a net profit of $$ 10$ for each of the first 1000 units produced each week. For each unit over 1000 produced, there is 2 cents less profit per unit. How many units should be produced each week to net the greatest profit?

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