Problem 29 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

If the function \(m(t)=\frac{1}{t+2}\) and \(h(t)=t-2\), then is it ever true that \(m(h(t))=h(m(t))\) ?

Step-by-Step Solution

Answer

No, the statement \(m(h(t))=h(m(t))\) does not hold true.

Step 1: Substitute h(t) into m(t)

Start by substituting \(h(t)\) into \(m(t)\) to get \(m(h(t))\). Therefore, \[ m(h(t)) = m(t-2)= \frac{1}{(t-2)+2} = \frac{1}{t} \]

Step 2: Substitute m(t) into h(t)

Next, substitute \(m(t)\) into \(h(t)\) to get \(h(m(t))\). Therefore, \[ h(m(t))= h\left(\frac{1}{t+2}\right) = \frac{1}{t+2}-2 = \frac{1-2(t+2)}{t+2} = \frac{1-2t-4}{t+2} = \frac{-2t-3}{t+2} \]

Step 3: Compare the results

Finally, compare \(m(h(t))\) with \(h(m(t))\). From the obtained results, it is seen that: \( m(h(t))=\frac{1}{t} \neq \frac{-2t-3}{t+2} = h(m(t)) \] Hence it is clear that the two resultant functions are not the same, disproving the original assertion.

Key Concepts

Inverse Functions

Inverse functions are an important concept in mathematics that allows us to reverse the effect of a given function. In simple terms, if you have a function and you also know its inverse, you can return to the original value after applying the inverse. Suppose we have a function \( f(x) \), its inverse is typically written as \( f^{-1}(x) \). For a function to have an inverse, it must be both one-to-one (injective) and onto (surjective). These conditions ensure that every input has a unique output.
In the context of our exercise, if we were looking to find an inverse relationship, we would need a setting where applying a function followed by its inverse would bring us back to our starting point. However, in this problem, we're dealing with two different functions, \( m(t) \) and \( h(t) \), rather than a function and its inverse. So, we're not really focusing on inverse functions here, but understanding this concept helps clarify why \( m(h(t)) \) and \( h(m(t)) \) might not be equal, as inverses are not involved.

Rational Functions

Rational functions are functions expressed as the quotient of two polynomials, where the numerator and the denominator are both polynomials. A simple example is \( \frac{1}{x+2} \), which is a rational function because it divides the constant 1 by the polynomial \((x+2)\).
In the exercise, the function \( m(t) = \frac{1}{t+2} \) is a rational function. One key property of rational functions is their restrictions. Because they feature division by a polynomial, they are undefined where the denominator equals zero. In \( m(t) \), it's crucial to remember that \( t eq -2 \) because that would make the denominator zero, which is undefined.
Rational functions are vital in this problem because when \( h(t) \) is composed with \( m(t) \), it doesn't result in a rational function equal to when \( m(t) \) is composed with \( h(t) \). This non-equivalence helps illustrate the importance of understanding restrictions and behavior of each composition.

Function Equality

Function equality is a concept that determines when two functions produce the same outputs for all input values. For two functions \( f(x) \) and \( g(x) \) to be equal, \( f(x) = g(x) \) for every \( x \) in the domain where both functions are defined.
In our math problem, we are asked if \( m(h(t)) = h(m(t)) \). Through our computations in the step-by-step solution, we derived that \( m(h(t)) = \frac{1}{t} \) while \( h(m(t)) = \frac{-2t-3}{t+2} \). These expressions are clearly not equal, as they produce different outputs for values of \( t \) within their respective domains.
Understanding function equality is crucial. It requires not just solving the expressions but also analyzing their outputs over valid inputs to ascertain equivalence. Differences in domains, forms, or values will indicate non-equality, as demonstrated in this exercise where composition order impacts the results.