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Chapter 7: Problem 25

Show that the equation \(e^{x}=1+x+\frac{x^{2}}{2}\) has a real root in \([-1,1]\).

Short Answer

Expert verified
The equation \(e^{x} = 1+x+\frac{x^{2}}{2}\) has a real root in \([-1,1]\) because the function \(e^{x}-1-x-\frac{x^{2}}{2}\) is continuous on \([-1,1]\) and takes on positive and negative values within this interval. By the Intermediate Value Theorem, it therefore has a root within this interval.

Step by step solution

01

Set the Function

Define the function \(f(x) = e^{x} - 1 - x - \frac{x^{2}}{2}\). We are interested in finding the roots of this function, which translate to the solutions of the initial equation.
02

Calculate the Endpoints

Calculate the values of the function at the endpoints of the interval \([-1,1]\). Specifically, find \(f(-1)\) and \(f(1)\). We find that \(f(-1) = e^{-1} - 1 - (-1) - \frac{1}{2} > 0\) and \(f(1) = e^{1} - 1 - 1 - \frac{1}{2} < 0\).
03

Apply the Intermediate Value Theorem

Use the Intermediate Value Theorem (IVT), which applies to continuous functions. Since \(f(x)\) is continuous on \([-1,1]\) and \(f(1)<0<f(-1)\), there must be a real number \(c\) in \([-1,1]\) such that \(f(c) = 0\), by the IVT. This shows that the equation \(e^{x} = 1 + x + \frac{x^{2}}{2}\) has a root in \([-1,1]\).

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Key Concepts

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

Real Roots
In algebra, the term "real roots" refers to the real number solutions where a given equation equals zero. When we find a real root of a function, like in the equation \[ e^{x} = 1 + x + \frac{x^{2}}{2} \], we are identifying the values of \( x \) that make the expression equal to zero. Real roots are the points where the graph of the function intersects the x-axis.

Real roots are important because they represent solutions in practical problems. Whether it鈥檚 in physics, engineering, or economics, finding where a mathematical model hits zero often signifies a meaningful result.
  • Real roots are solutions to equations set to zero.
  • They represent intersections with the x-axis on a graph.
  • Finding real roots is essential in interpreting many natural phenomena.
Continuous Functions
A function is said to be continuous if you can draw it without lifting your pencil from the paper. More formally, a continuous function has no breaks, jumps, or holes at any point in its domain.

The function involved in the problem, \( f(x) = e^{x} - 1 - x - \frac{x^{2}}{2} \), is continuous over the interval \([-1, 1]\). This continuity is crucial because it allows us to use the Intermediate Value Theorem (IVT) to find roots.
  • Continuity means no jumps or breaks in a function.
  • Continuous functions make it possible to apply the Intermediate Value Theorem.
  • Checking for continuity is a vital step in root-finding procedures.
Exponential Functions
Exponential functions are powerful mathematical expressions of the form \( a^{x} \), where a is a constant, and x is the exponent. The function \( e^{x} \) is a standard example, where \( e \) is Euler's number (approximately 2.71828).

In our equation \[ e^{x} = 1 + x + \frac{x^{2}}{2} \], an exponential function is balanced against a polynomial expression on the right side. Exponential functions differ from polynomials as they can grow or decay rapidly, depending on the sign of x, and significantly influence the behavior of the equation.
  • Exponential functions have the form \( a^{x} \).
  • They exhibit rapid growth or decay, based on the exponent's sign.
  • \( e^{x} \) is a common exponential function used in mathematics.
Polynomial Equations
Polynomial equations consist of sums and differences of terms, each being a constant multiplied by a variable raised to a whole number power. They appear frequently in various forms and complexities.

In the exercise, the right side of our equation \[ e^{x} = 1 + x + \frac{x^{2}}{2} \] represents a simple polynomial. The polynomial here is quadratic, characterized by terms with powers up to two. It interacts with the exponential function \( e^{x} \) to form a mixed equation.

Polynomials are simpler and have different properties compared to exponential functions, which can make them more predictable over certain intervals.
  • Polynomial equations have terms with variables raised to integer powers.
  • They can be simple (linear, quadratic) or complex (higher degrees).
  • Understanding polynomials helps in analyzing and predicting equation behaviors.

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Most popular questions from this chapter

Find the interval of the increases or decreases for the function \(f(x)=\int_{-1}^{x}\left(t^{2}+2 t\right)\left(t^{2}-1\right) d t\)

If \(f(x)=2 x+\cot ^{-1} x+\left(\sqrt{1+x^{2}}+1\right)\), then \(f(x)\) (a) inc. on \([0, \infty)\) (b) dec. on \([0, \infty)\) (c) neither inc. nor dec. on \([0, \infty)\) (d) inc. on \((-\infty, \infty)\)

Let \(f(x)=\sin ^{2} x-(2 a+1) \sin x+(a-c) .\) If \(f(x) \leq 0\) for all \(x\) in \(\left[0, \frac{\pi}{2}\right]\), then the range of \(a\) is (a) \([-3,0]\) (b) \([3, \infty)\) (c) \([-3,3]\) (d) \((-\infty, 3]\).

Let \(f(x)=x^{3}+6 x^{2}+p x+2 .\) If the largest possible interval in which \(f(x)\) is dec. function in \((-3,-1)\) then \(p\) is (a) 2 (b) 6 (c) 8 (d) 9

The function \(f(x)=\frac{1}{x}\) on its domain is (a) increasing (b) decreasing (c) constant (d) none
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