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Multiple Choice What is the linearization of \(f(x)=e^{x}\) at \(x=1 ?\) (A) \(y=e \quad\) (B) \(y=e x \quad\) (C) \(y=e^{x}\) \((\mathbf{D}) y=x-e \quad\) (\mathbf{E} ) ~ \(y=e(x-1)\)

Short Answer

Expert verified
The linearization of \(f(x)=e^{x}\) at \(x=1\) is \(y = e(x - 1)\)

Step by step solution

01

Find the Derivative

The first step is to find the derivative of the function \(f(x) = e^{x}\). Since the derivative of \(e^{x}\) is \(e^{x}\), the derivative of the function at \(x=1 (f'(1))\) is \(e^{1}\), which equals \(e\).
02

Create the Linearization Formula

With the derivative in hand, the linearization \(L(x)\) of the function \(f(x) = e^{x}\) at a point \(x=a\) is given by the equation \(L(x) = f(a) + f'(a)(x-a)\). Substitute \(a=1\), \(f(1)=e^{1}=e\), and \(f'(1)=e\) into the linearization equation to get \(L(x) = e + e(x - 1)\). This equals \(ex - e\), which can also be written as \(y = e(x - 1)\).
03

Compare to Given Options

Comparing the linear function obtained in the previous step to the answer options in the question, it is clear that the correct answer is (E) \(y = e(x - 1)\)

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Key Concepts

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

Differentiation
Understanding the process of differentiation is crucial in identifying how a function changes at any given point. Differentiation is the action of computing a derivative. The derivative of a function provides us with the rate at which the function's value is changing at any point. For instance, if we're dealing with acceleration in physics, the process can tell us how quickly the velocity of an object is changing over time.

In mathematical terms, given a function f(x), the derivative f'(x) is the limit of the average rate of change of the function over an interval, as the interval's length approaches zero. Applying this to exponential functions like e^x, where e is Euler's number, the derivative yields the same exponential function, that is, f'(x) = e^x. In the context of the provided exercise, when differentiating f(x) = e^x, we get f'(x) = e^x, which at x=1 gives us f'(1) = e.

Expanding the knowledge of differentiation not only helps us solve textbook exercises but also enhances our understanding of the dynamics of various natural and scientific phenomena.
Tangent Line Approximation
The tangent line approximation, also known as linearization, is a method for approximating the value of a function near a particular point using the tangent line at that point. This concept is extremely helpful in situations where the function itself is too complex for straightforward evaluation.

For a differentiable function f(x), the tangent line at x=a can be represented as L(x) = f(a) + f'(a)(x - a). Here, f(a) is the value of the function at x=a, and f'(a) is the value of the derivative (or the slope of the tangent line) at the same point. This linear equation provides a simple approximation for f(x) when x is near a, making it a powerful tool for estimations and simplifying complex expressions.

Applying this to the example in the exercise, at x=1 for the function f(x) = e^x, we would obtain the linear approximation L(x) = e + e(x - 1), which simplifies to y = e(x - 1), matching answer option (E). This approximation is particularly useful for quick calculations and insights into the behavior of the function near the point of tangency.
Exponential Functions
An exponential function is one of the most important concepts in mathematics, characterized by a constant base raised to a variable exponent. The general form is f(x) = b^x where b is a positive real number. Among these functions, the most notable is e^x, where e (approximately 2.71828) is Euler's number, a fundamental constant in mathematics.

The special property of e^x is that it is its own derivative, meaning that the rate of change of this function at any point is equal to the value of the function at that point. This property makes e^x incredibly important in various scientific calculations, including growth processes, radioactive decay, and compound interest in finance.

In the context of the exercise, recognizing that f(x) = e^x is an exponential function allows us to apply the concepts of differentiation and tangent line approximation effectively to find the linearization at x=1. Understanding the unique characteristics of exponential functions can greatly enhance one's ability to solve a wide range of real-world problems.

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Most popular questions from this chapter

In Exercises 62 and \(63,\) feel free to use a CAS (computer algebra system), if you have one, to solve the problem. Logistic Functions Let \(f(x)=c /\left(1+a e^{-h x}\right)\) with \(a>0\) \(a b c \neq 0\) (a) Show that \(f\) is increasing on the interval \((-\infty, \infty)\) if \(a b c>0\) and decreasing if \(a b c<0\) . (b) Show that the point of inflection of \(f\) occurs at \(x=(\ln |a|) / b\)

The Linearization is the Best Linear Approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c(m\) and \(c\) constants). If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions i. \(E(a)=0\) ii. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \((x-a)\) .

Strength of a Beam The strength S of a rectangular wooden beam is proportional to its width times the square of its depth. (a) Find the dimensions of the strongest beam that can be cut from a 12-in. diameter cylindrical log. (b) Writing to Learn Graph \(S\) as a function of the beam's width \(w,\) assuming the proportionality constant to be \(k=1 .\) Reconcile what you see with your answer in part (a). (c) Writing to Learn On the same screen, graph \(S\) as a function of the beam's depth \(d,\) again taking \(k=1 .\) Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of \(k ?\) Try it.

Production Level Show that if \(r(x)=6 x\) and \(c(x)=\) \(x^{3}-6 x^{2}+15 x\) are your revenue and cost functions, then the best you can do is break even (have revenue equal cost).

True or False A continuous function on a closed interval must attain a maximum value on that interval. Justify your answer.

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