Question

In applications, the symbols used for the independent and dependent variables are often based on common usage. So, rather than using \(y=f(x)\) to represent a function, an applied problem might use \(C=C(q)\) to represent the cost C of manufacturing qunits of a good. Because of this, the inverse notation \(f^{-1}\) used in a pure mathematics problem is not used when finding inverses of applied problems. Rather, the inverse of a function such as \(C=C(q)\) will be \(q=q(C) .\) So \(C=C(q)\) is a function that represents the cost \(\mathrm{C}\) as a function of the number \(q\) of units manufactured, and \(q=q(C)\) is a function that represents the number \(q\) as a function of the cost \(C\). Illustrate this idea. Gravity on Earth If a rock falls from a height of 100 meters above Earth, the height \(H\) (in meters) after \(t\) seconds is approximately $$ H(t)=100-4.9 t^{2} $$ (a) In general, quadratic functions are not one-to-one. However, the function \(H\) is one-to-one. Why? (b) Find the inverse of \(H\) and verify your result. (c) How long will it take a rock to fall 80 meters?

Short answer

a) The height function \(H(t) \) is one-to-one because it decreases monotonically for \(t \geq 0\). b) The inverse is \(t = \sqrt{\frac{100-H}{4.9}} \).c) It takes approximately 4.04 seconds to fall 80 meters.

Step-by-step solution

- Understanding the one-to-one behavior (Part a)

To determine if the function is one-to-one, understand that for a function to be one-to-one, each value of the independent variable must map to a unique value of the dependent variable. Given is the height function \(H(t) = 100 - 4.9t^2\).Since it models the height of a freely falling rock, it only decreases with time from a maximum initial value of 100 meters (at time t=0) as \(t\) increases. Therefore, it’s a decreasing function for \(t \geq 0\) and cannot have the same height at two different instances. Thus, it is a one-to-one function.

- Finding the inverse function (Part b)

To find the inverse function, solve for \(t\) in terms of \(H\).**1.** Start with the given function:\[ H = 100 - 4.9t^2 \]**2.** Isolate the term involving \(t\):\[ 100 - H = 4.9t^2 \]**3.** Solve for \(t^2\):\[ t^2 = \frac{100 - H}{4.9} \]**4.** Take the square root of both sides (consider positive root only as time cannot be negative):\[ t = \sqrt{\frac{100 - H}{4.9}} \]Therefore, the inverse function is: \[ t = \sqrt{\frac{100 - H}{4.9}} \]

- Verifying the inverse

To verify the inverse, check if applying the original function and its inverse in succession returns the initial input value.**1.** Start with the original function and apply the inverse:\[ H(\sqrt{\frac{100 - H}{4.9}}) \]**2.** Substitute \(t = \sqrt{\frac{100 - H}{4.9}}\):\[ H = 100 - 4.9(\sqrt{\frac{100 - H}{4.9}})^2 \]\[ H = 100 - (100 - H) \]\[ H = H \] This confirms that the inverse function is correct.

- Calculating time for a given height (Part c)

To find how long it takes for the rock to fall 80 meters (i.e., height = 100 - 80 = 20 meters), use the inverse function:\[ t = \sqrt{\frac{100 - 20}{4.9}} \]Calculate the terms inside the square root first:\[ t = \sqrt{\frac{80}{4.9}} \]\[ t \approx \sqrt{16.33} \]Find the square root:\[ t \approx 4.04 seconds \]

Key concepts

One-to-One Functions

A function is called a *one-to-one function* if every value of the independent variable results in a unique value of the dependent variable. This means there are no two different inputs which yield the same output.
To understand if a function is one-to-one, we can use the horizontal line test: if any horizontal line intercepts the graph of the function at most once, the function is one-to-one.
In the given problem, the height function is specified as \( H(t) = 100 - 4.9t^2 \). Here’s why \(H(t)\) is one-to-one:

• It models the height of a rock falling from 100 meters.
• As time increases, the height decreases continuously from 100 meters at \(t = 0\), making it a strictly decreasing function.

Since it’s always decreasing after \(t = 0\), no two different times \(t\) can give the same height \(H\), confirming that the function is indeed one-to-one.

Quadratic Functions

Quadratic functions are polynomial functions of the form \( f(x) = ax^2 + bx + c\) where \(a, b\), and \(c\) are constants and \(a eq 0\).
These functions typically produce a parabolic graph that opens upwards if \(a > 0\) or downwards if \(a < 0\).
In the exercise, the function \(H(t) = 100 - 4.9t^2\) is a specific type of quadratic function. It has the structure of \(ax^2 + bx + c\) where:

• The coefficient \(a=-4.9\) is negative, indicating that the parabola opens downwards.
• The term \(bx\) is absent, implying \(b=0\).
• The constant \(c=100\) shifts the vertex of the parabola to \(t=0, H=100\).

Because of the downward opening and absence of \(b\), this function strictly decreases over time \(t \geq 0\), making it useful in modeling the height of a free-falling rock.

Finding Inverse

Finding the inverse of a function essentially means swapping the roles of the independent and dependent variables, and then solving for the original independent variable in terms of the new dependent variable.
To find the inverse of the function \(H(t) = 100 - 4.9t^2\):

• Start with the equation \( H(t) = 100 - 4.9t^2 \).
• Isolate \(t\): \(100 - H = 4.9t^2\)
• Solve for \(t^2\): \(t^2 = \frac{100 - H}{4.9}\)
• Recognize that since \(t\) represents time and cannot be negative, take the positive square root: \( t = \sqrt{\frac{100 - H}{4.9}}\)

Hence, the inverse function is: \( t = \sqrt{\frac{100 - H}{4.9}}\). To verify the inverse, you can substitute back and ensure that applying the inverse function to the original results returns the initial input value.
Finally, to use the inverse function practically, like determining how long it takes for a rock to fall to a certain height, apply the given height \(H\) into the inverse function and solve for \(t\). For example, if the height is 20 meters, use:
\( t = \sqrt{\frac{100 - 20}{4.9}} \approx 4.04 \) seconds.