Problem 44 - Contemporary Precalculus

Question

None of the functions has an inverse. State at least one way of restricting the domain of the function (that is, find a function with the same rule and a smaller domain so that the restricted function has an inverse. Then find the rule of the inverse function. Example: \(f(x)=x^{2}\) has no inverse. But the function \(h\) with domain all \(x \geq 0\) and rule \(h(x)=x^{2}\) is increasing (its graph is the right half of the graph of \(f-\) see Figure \(2-2\) on page 78 )and therefore has an inverse. $$f(x)=3(x+5)^{2}+2$$

Step-by-Step Solution

Answer

Answer: The inverse function of the restricted function \(h(x)\) is \(h^{-1}(x) = \sqrt{\frac{x-2}{3}} - 5\).

Step 1: Identify the function and its properties

We are given the function $$f(x) = 3(x + 5)^2 + 2$$ This is a quadratic function, which means it is not one-to-one and onto as is.

Step 2: Restrict the domain

To create a one-to-one and onto function from the given function, we can consider only the increasing or decreasing part of the function. Since the function has the form \(ax^2 + bx + c\), with \(a > 0\), we know that it will open upwards, so it increases as we move to the right of the vertex. Therefore, we choose the domain as all \(x\) greater than or equal to the x-coordinate of the vertex. To find the vertex, we can use the formula \(x = -\frac{b}{2a}\) for the x-coordinate. However, in our given function, it is clear that the vertex is at \((-5, 2)\). Therefore, we restrict the function \(h : [-5, \infty) \to \mathbb{R}\) with rule, $$h(x)=3(x+5)^2+2$$

Step 3: Find the inverse function

Now, let's find the inverse of the restricted function \(h(x)\). To do this, swap \(x\) and \(y\) in the equation and then solve for \(y\). $$x = 3(y+5)^2 + 2$$ Perform the following steps to isolate \(y\): 1. Subtract 2 from both sides. $$x-2 = 3(y+5)^2$$ 2. Divide by 3. $$\frac{x-2}{3} = (y+5)^2$$ 3. Take the square root of both sides. Since we have restricted the domain to \(x \geq -5\), we only need to consider the positive square root. $$\sqrt{\frac{x-2}{3}} = y + 5$$ 4. Subtract 5. $$h^{-1}(x) = \sqrt{\frac{x-2}{3}} - 5$$ Now we have found the inverse function of the restricted function \(h(x)\): $$h^{-1}(x) = \sqrt{\frac{x-2}{3}} - 5$$

Key Concepts

Quadratic Function

At the heart of this exercise is the quadratic function. Quadratic functions are mathematical expressions in the form of \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). They are called quadratic because the highest power of the variable \( x \) is square. These functions create a u-shaped curve called a parabola when graphed. The vertex of the parabola is either its highest or lowest point, depending on whether the parabola opens upwards or downwards. Quadratic functions are not one-to-one. This means that for some value of the output, more than one input might fit, which typically happens because the parabola is symmetrical. In this exercise, by restricting the domain of the function, it is possible to make it one-to-one. This helps in defining an inverse function which will be addressed further.

Domain Restriction

Domain restriction is an essential concept for transforming certain functions so they can have inverses. For a function to have an inverse, it must be bijective

• Injective: Different inputs should produce different outputs
• Surjective: Every possible output should be matched by an input

Quadratic functions naturally fail the injective condition as discussed. To solve this, we limit or "restrict" the domain to focus on one half of the parabola. For example, focusing only on the part where the function is consistently increasing or decreasing can achieve this.In the given exercise, the restriction was set to \( x \geq -5 \), which focuses on the increasing part of the quadratic function. By restricting the domain this way, we ensure that for each \( y \) value, there is only one \( x \) value, hence making it one-to-one and allowing us to find its inverse.

One-to-One Function

A function is one-to-one when each input maps to exactly one unique output. This property is crucial because only one-to-one functions have inverses. When a function faithfully represents an input-output relationship without duplicates, it's considered one-to-one. With quadratic functions, due to their symmetry, one-to-one mapping isn't naturally present unless restricted to a specific domain where the function is strictly increasing or decreasing. This step is vital to enable finding an inverse.In the case of \( f(x) = 3(x + 5)^2 + 2 \), by restricting to the domain \([-5, \infty)\), the parabola can be seen as an increasing function. Consequently, it's one-to-one over this restricted domain, and this allows an inverse to be calculated for the specified range of inputs. An inverse function then swaps the roles of inputs and outputs, so for \( h(x) \), the inverse \( h^{-1}(x) \) is derived by reversing the processes applied to \( x \).