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Chapter 4: Problem 56

Show that the function \(T(x)=60 D(60+x)^{-1}\) gives the time in minutes required to drive \(D\) miles at \(60+x\) miles per hour.

Short Answer

Expert verified
Answer: Yes, the function \(T(x) = 60 D(60+x)^{-1}\) represents the time in minutes required to drive \(D\) miles at a speed of \(60+x\) miles per hour.

Step by step solution

01

Understand the given function

The given function is \(T(x) = 60 D(60+x)^{-1}\). Here, \(T(x)\) represents the time in minutes required to drive \(D\) miles, and \(x\) represents the additional speed (in miles per hour) above the base speed of \(60\) miles per hour.
02

Rewrite the function using the speed, distance, and time formula

We know that speed = \(\frac{distance}{time}\). Let's denote the speed as \(S\). Since we are given the function in terms of time, we need to rewrite it so that we can express \(S\) in terms of \(T(x)\) and \(D\). We have \(S = 60+x\). Substituting this into the speed formula, we get: $$S = \frac{D}{T(x)}$$
03

Replace the speed with the given expression for S

Now, we will replace \(S\) with the given expression, \(60+x\): $$60+x = \frac{D}{T(x)}$$
04

Isolate T(x) to show the given function

We want to isolate \(T(x)\) to show that the given function represents the correct time. To do this, multiply both sides of the equation by \(T(x)\): $$(60+x)T(x) = D$$ Next, divide both sides by \((60+x)\) to isolate \(T(x)\): $$T(x) = \frac{D}{60+x}$$
05

Compare the given function with the derived function

Comparing the derived function to the given function, we can see that they are equivalent: Given function: \(T(x) = 60 D(60+x)^{-1}\) Derived function: \(T(x) = \frac{D}{60+x}\) Since both functions are equal, this shows that the function \(T(x) = 60 D(60+x)^{-1}\) gives the time in minutes required to drive \(D\) miles at \(60+x\) miles per hour.

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