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Chapter 3: Problem 2

What is the tangent line approximation to \(e^{x}\) near \(x=0 ?\)

Short Answer

Expert verified
The tangent line approximation to \( e^{x} \) near \( x=0 \) is \( y = 1 + x \).

Step by step solution

01

Understand the Problem

We are asked to find the tangent line approximation to the function \( e^{x} \) at \( x=0 \). This means we need to find the line that just "touches" the curve of \( e^{x} \) at the point where \( x = 0 \).
02

Recall the Tangent Line Formula

The formula for the tangent line to a function \( f(x) \) at a point \( x=a \) is given by \( y = f(a) + f'(a)(x-a) \).
03

Find the Value of the Function at the Point

First, find the value of the function \( e^{x} \) at \( x = 0 \). Evaluate \( e^{0} = 1 \).
04

Find the Derivative of the Function

The derivative of \( e^{x} \), \( f'(x) \), is \( e^{x} \) itself because the derivative of \( e^{x} \) is \( e^{x} \).
05

Evaluate the Derivative at the Point

Evaluate the derivative at \( x = 0 \). Therefore, \( f'(0) = e^{0} = 1 \).
06

Calculate the Tangent Line Equation

Plug the values into the tangent line formula: \( y = e^{0} + e^{0}(x-0) = 1 + 1 \cdot x = 1 + x \). Thus, the equation of the tangent line is \( y = 1 + x \).

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Key Concepts

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

Derivative
A derivative is an essential part of calculus, representing the rate at which a function is changing at any given point. You can think of it as the function's slope at that specific point. In mathematical terms, the derivative is defined as:
  • If you have a function \( f(x) \), its derivative \( f'(x) \) tells how \( y \) (the output) changes as \( x \) (the input) changes.
  • The derivative provides the slope of the tangent line to the graph of the function at any point \( x \).
For the exponential function \( e^x \), the amazing aspect is that its derivative is itself, meaning \( f'(x) = e^x \). This is crucial because, for functions like \( e^x \), computing derivatives simplifies dramatically, which is helpful in many real-life contexts where exponential growth comes into play.
Function Evaluation
Function evaluation involves determining the output of a mathematical function given a certain input. It's like seeing what comes out when you put something into a machine. For instance:
  • If you have a machine \( e^x \) and you put in 0, you observe the output.
  • At \( x = 0 \), especially for the exponential function, this evaluation gives you \( e^0 \).
The nice thing about \( e^x \) is that any number raised to the power of zero is 1, so \( e^0 = 1 \). This result is used further when calculating your tangent line approximation, as it's crucial for figuring out where the tangent line touches the curve.
Exponential Function
Exponential functions are powerful mathematical tools used to describe situations of rapid increase or decrease. This type of function is written as \( e^x \) where \( e \) is approximately 2.71828, a mathematical constant. This exponent represents continuous growth or decay:
  • Commonly seen in real-world scenarios like population growth, radioactive decay, and financial investments.
  • In calculus, the characteristic property is its unchanged derivative, i.e., \( \frac{d}{dx}(e^x) = e^x \).
  • This unchanged nature simplifies solving differential equations, which describe changes over time.
Recognizing the exponential function and understanding its properties helps in many areas of mathematics and science, and is fundamental when working with tangent line approximations.
Tangent Line Formula
The tangent line formula is a significant concept when assessing how a curve behaves at a particular point. It's akin to finding a flat approximation of a curve, right where the curve narrowly passes:
  • The equation of the tangent line is derived from \( y = f(a) + f'(a)(x-a) \).
  • Here, \( f(a) \) provides the height of the curve, while \( f'(a) \) uses the slope.
  • The term \((x-a)\) conveys the x-distance from the point where we're forming the tangent.
We apply this formula by plugging in numbers for \( e^x \) at \( x=0 \), obtaining \( y = 1 + x \). This linear equation offers a close approximation of the curve precisely at \( x=0 \), bridging the gap between curves and straight lines.

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Most popular questions from this chapter

Give an example of: An interval where the Mean Value Theorem does not apply when \(f(x)=1 / x\)

(a) Show that \(1-x\) is the local linearization of \(\frac{1}{1+x}\) near \(x=0\) (b) From your answer to part (a), show that near \(x=0$$$\frac{1}{1+x^{2}} \approx 1-x^{2}$$ (c) Without differentiating, what do you think the derivative of \)\frac{1}{1+x^{2}}\( is at \)x=0 ?$

Find a formula for the error \(E(x)\) in the tangent line approximation to the function near \(x=a\). Using a table of values for \(E(x) /(x-a)\) near \(x=a\), find a value of \(k\) such that \(E(x) /(x-a) \approx k(x-a) .\) Check that, approximately, \(k=f^{\prime \prime}(a) / 2\) and that \(E(x) \approx\left(f^{\prime \prime}(a) / 2\right)(x-a)^{2}\). $$f(x)=e^{x}, a=0$$

Are the statements true or false? Give an explanation for your answer. The graph of \(\ln \left(x^{2}\right)\) is concave up for \(x>0\).

Describe and explain the behavior of \(\sinh x\) as \(x \rightarrow \infty\) and as \(x \rightarrow-\infty\)
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