Graph the equation as two separate functions. Set the left side of the equation equal to [tex]f(x)[/tex] and the right side of the equation equal to [tex]g(x)[/tex].
[tex](x+1)^2-2=\frac{2}{x}[/tex]
Use the graph to answer these questions:
- When does [tex]f(x)=g(x)[/tex] ?
- What are the [tex]x[/tex]-values of the solution?
- How do the solutions compare with the solutions when solving the equation algebraically?
Answer (1)
The problem requires graphing f ( x ) = ( x + 1 ) 2 − 2 and g ( x ) = x 2 and finding where f ( x ) = g ( x ) .
Algebraically, the equation ( x + 1 ) 2 − 2 = x 2 is transformed into x 3 + 2 x 2 − x − 2 = 0 .
Factoring the cubic equation yields ( x − 1 ) ( x + 1 ) ( x + 2 ) = 0 , giving solutions x = 1 , − 1 , − 2 .
The solutions are the x -values where the two functions intersect, which are x = 1 , − 1 , − 2 .
Explanation
Understanding the Problem We are given the equation ( x + 1 ) 2 − 2 = x 2 . We need to graph the two functions f ( x ) = ( x + 1 ) 2 − 2 and g ( x ) = x 2 and find the x -values where they intersect. These x -values are the solutions to the given equation.
Defining the Functions First, let's expand f ( x ) : f ( x ) = ( x + 1 ) 2 − 2 = x 2 + 2 x + 1 − 2 = x 2 + 2 x − 1 and g ( x ) = x 2 We will graph these two functions.
Graphical Solution The points where f ( x ) = g ( x ) are the solutions to the equation. By graphing the two functions, we can visually identify the intersection points. The x -coordinates of these points are the solutions.
Algebraic Manipulation Now, let's solve the equation algebraically: ( x + 1 ) 2 − 2 = x 2 x 2 + 2 x − 1 = x 2 Multiply both sides by x :
x ( x 2 + 2 x − 1 ) = 2 x 3 + 2 x 2 − x = 2 x 3 + 2 x 2 − x − 2 = 0 We need to find the roots of the cubic equation x 3 + 2 x 2 − x − 2 = 0 .
Solving the Cubic Equation We can factor the cubic equation by grouping: x 3 + 2 x 2 − x − 2 = 0 x 2 ( x + 2 ) − 1 ( x + 2 ) = 0 ( x 2 − 1 ) ( x + 2 ) = 0 ( x − 1 ) ( x + 1 ) ( x + 2 ) = 0 The solutions are x = 1 , x = − 1 , and x = − 2 .
Checking for Extraneous Solutions However, we must check for extraneous solutions since we multiplied by x in an earlier step. If x = 0 , then g ( x ) is undefined, so x cannot be 0. The solutions x = 1 , x = − 1 , and x = − 2 are all valid since none of them make the denominator of g ( x ) equal to zero. Therefore, the solutions are x = 1 , x = − 1 , and x = − 2 .
Comparing Solutions The solutions obtained graphically should match the solutions obtained algebraically. The x -values where f ( x ) = g ( x ) are x = 1 , x = − 1 , and x = − 2 .
Examples
Understanding the points where two functions intersect is crucial in many real-world applications. For instance, in economics, the intersection of supply and demand curves determines the market equilibrium price and quantity. Similarly, in physics, finding the points where two trajectories intersect can help predict collisions. Graphing functions and solving equations algebraically are fundamental tools for analyzing and solving such problems.
