Question

a. Show that \(f(x)=-x^{2}+x+1\) on \(\left[\frac{1}{2}, \infty\right)\) and \(g(x)=\frac{1}{2}+\sqrt{\frac{5}{4}-x}\) on \(\left(-\infty, \frac{5}{4}\right)\) are inverses of each other. b. Solve the equation \(-x^{2}+x+1=\frac{1}{2}+\sqrt{\frac{5}{4}-x}\). Hint: Use the result of part (a).

Short answer

In this problem, it was demonstrated that the functions \(f(x) = -x^2 + x + 1\) and \(g(x) = \frac{1}{2} + \sqrt{\frac{5}{4} - x}\) are inverses of each other on their respective domains, as both compositions \(f(g(x))\) and \(g(f(x))\) equal \(x\). The solution of the given equation \(-x^2 + x + 1=\frac{1}{2}+\sqrt{\frac{5}{4}-x}\) using the result from part (a) shows that it has a single valid intersection point at \(x = \frac{2 + \sqrt{2}}{2}\).

Step-by-step solution

Compute the composition f(g(x))

First, we compute the composition \(f(g(x))\): \(f(g(x)) = f( \frac{1}{2} + \sqrt{\frac{5}{4} - x} ) = -(\frac{1}{2} + \sqrt{\frac{5}{4} - x})^2 + (\frac{1}{2} + \sqrt{\frac{5}{4} - x}) + 1\)

Simplify the expression f(g(x))

Now, we simplify the expression for \(f(g(x))\): \(-(\frac{1}{2} + \sqrt{\frac{5}{4} - x})^2 + (\frac{1}{2} + \sqrt{\frac{5}{4} - x}) + 1 = -(\frac{1}{4} + x + \sqrt{(\frac{5}{4}-x)}) + (\frac{1}{2} + \sqrt{\frac{5}{4} - x}) + 1 = x\)

Compute the composition g(f(x))

Next, we compute the composition \(g(f(x))\): \(g(f(x)) = g(-x^2 + x + 1) = \frac{1}{2} + \sqrt{\frac{5}{4} - (-x^2 + x + 1)}\)

Simplify the expression g(f(x))

Now, we simplify the expression for \(g(f(x))\): \(\frac{1}{2} + \sqrt{\frac{5}{4} - (-x^2 + x + 1)} = \frac{1}{2} + \sqrt{x^2 - x + \frac{1}{4}} = x\) Since both compositions \(f(g(x)) = x\) and \(g(f(x)) = x\), these two functions are inverses of each other.

Use the result of part (a) to solve the equation

From part (a), we know that \(f(x)\) and \(g(x)\) are inverses of each other. Therefore, we can rewrite the given equation as follows: \(f(x) = g(x)\) Since the two functions are inverses, this means that they intersect at a single point. Therefore, their compositions will be equal to the identity function at this intersection point. We already showed that \(f(g(x)) = x\). Therefore, at the intersection point, the x-coordinate will be equal to the output of the function. Thus, we can find the intersection point by solving for \(x\): \(x = \frac{1}{2} + \sqrt{\frac{5}{4} - x}\)

Solve for x in the equation

To solve for \(x\), we can rewrite the equation as follows: \(\sqrt{\frac{5}{4} - x} = x - \frac{1}{2}\) Squaring both sides, we get: \(\frac{5}{4} - x = (x - \frac{1}{2})^2\) Expanding the right side, \(\frac{5}{4} - x = x^2 - x + \frac{1}{4}\) Now, we can solve for \(x\): \(x^2 - 2x + \frac{1}{2} = 0\) This is a quadratic equation. We can use the quadratic formula to find the solution(s) for \(x\): \(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(\frac{1}{2})}}{2(1)}\) \(x = \frac{2 \pm \sqrt{4 - 2}}{2}\) \(x = \frac{2 \pm \sqrt{2}}{2}\) Since the domains of the functions are \(\left[\frac{1}{2}, \infty\right)\) and \(\left(-\infty, \frac{5}{4}\right)\), only one of the solutions is valid. We discard the solution \(x = \frac{2 - \sqrt{2}}{2}\) as it is not in the domain of both functions. Therefore, the valid solution for the intersection point is: \(x = \frac{2 + \sqrt{2}}{2}\) In conclusion, the two functions \(f(x) = -x^2 + x + 1\) and \(g(x) = \frac{1}{2} + \sqrt{\frac{5}{4} - x}\) are inverses of each other, and they intersect at the point \(\left(\frac{2 + \sqrt{2}}{2}, \frac{2 + \sqrt{2}}{2}\right)\).

Key concepts

Composition of Functions

When studying calculus, the composition of functions is a fundamental concept. A composition of two functions, denoted as f(g(x)), creates a new function, where g(x) is evaluated first and the result is then plugged into the function f(x). To verify if two functions are inverses of each other, one should check if the composition of one into the other yields the identity function, meaning f(g(x)) = x and g(f(x)) = x.

For example, if we have functions f(x) = -x^2 + x + 1 and g(x) = (1/2) + sqrt(5/4 - x), to prove they are inverses, we compose one into the other. If the result simplifies to x, this is a strong indicator that we indeed have a pair of inverse functions, provided that this holds true for both possible compositions f(g(x)) and g(f(x)).

Solving Quadratic Equations

Quadratic equations of the form ax^2 + bx + c = 0 are prevalent in algebra and can often appear in calculus problems, such as when finding the intersection point of functions. These equations are typically solved using methods like factoring, completing the square, or most universally, the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a).

In the context of our exercise, we encountered a quadratic equation while searching for the intersection point. By rearranging terms and applying the quadratic formula, we obtained the solutions for x. Remember to consider the domain of the functions in question when choosing the valid solution from the two possibilities provided by the formula.

Function Intersection Point

The intersection point of two functions is a set of coordinates where their graphs meet. For inverse functions, this is especially significant as they reflect across the line y = x and thus intersect at a point where their output equals the input value (x).

In our exercise, we used the fact that f(x) and g(x) are inverses to set up an equation, which when solved, gave us the intersection point of the functions. This point represents an x-value for which both functions yield the same result and essentially is where the graph of f intersects with g.

Domain of a Function

Understanding the Domain

The domain of a function is the complete set of possible values of the independent variable or variables for which the function is defined. This is critical when considering the validity of solutions as functions can only have outputs for input values within their domain.

In the discussed problem, we have specific domain restrictions for each function; f(x) is defined on [1/2, ∞) and g(x) on (-∞, 5/4). These restrictions guide us to dismiss any solutions to equations that fall outside these intervals. When solving for function intersection and inverse checks, keep in mind the domains to avoid incorporating invalid solutions.