91Ó°ÊÓ

A quadratic equation is a fundamental concept in algebra. It is any equation that can be re-arranged into the form:
\[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. In our specific case from the exercise, the equation took the form \( 2x^2 + x - 1 = 0 \).
Solving a quadratic equation involves finding values of \( x \) that make the equation true. The quadratic formula, a mainstay method, gives us a straightforward way to find these solutions:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here:

• \( b^2 - 4ac \) is known as the discriminant, which determines the nature of the roots (real or complex).
• If the discriminant is positive, as in our case where it was 9, there are two distinct real roots.
• For a quadratic like \( 2x^2 + x - 1 = 0 \), applying the formula gives us the solutions \( x = 0.5 \) and \( x = -1 \). Since only the positive solution is relevant to the problem scenario, we choose \( x = 0.5 \).

The cosine function, represented as \( \cos x \), is a fundamental trigonometric function that indicates the cosine of an angle within a right triangle or a unit circle context. It's repeating and periodic in nature with a cycle every \( 2\pi \) radians.
In many computational cases, especially within specific bounds or approximations, employing Taylor series expansions for trigonometric functions helps simplify problems. For example, the Taylor polynomial for \( \cos x \) at \( x=0 \), expressed as \( 1 - \frac{x^2}{2} \), approximates \( \cos x \) over a small range of \( x \).
This second-order Taylor polynomial captures the essence of the cosine near \( x=0 \) and is particularly useful for simplifying calculations without significant accuracy loss within a restricted interval, as was illustrated in our problem, replacing \( \cos x \) with \( 1 - \frac{x^2}{2} \). This approximation transformed our trigonometric equation into a quadratic one, easily solvable.

The concept of supply and demand is the backbone of economic theory, describing how the market reaches an equilibrium for goods and services.
In this exercise, demand was dictated by the function \( 4 \cos x \), and supply was given by \( x + 3 \) (in thousands), where \( x \) represents the passage of time in years.
The goal is to find the point where supply equals demand. Essentially, this is the market equilibrium point.

• The demand function reflects how much of a product consumers want at a given time.
• The supply function indicates how much of that product producers are willing to make available.
• By setting \( 4 \cos x \) equal to \( x + 3 \), we simulate finding when the quantity demanded by consumers precisely matches the quantity supplied by producers.

In the given problem, to find this equilibrium, we replaced the trigonometric \( \cos x \) function with its second Taylor polynomial to simplify our calculations, leading us to solve a quadratic equation.