In the given challenge, two different functions that model the speed of a car are given. The first function estimates the speed based on the braking distance, while the second function uses the power of the car to approximate its speed.

v1(d)v2(p)=30μd=838p

In these formulas, μ=0.6 is the coefficient of friction, d is the braking distance in feet, and p is the power in horsepower. Both speeds are measured in miles per hour.

a Find the domains D1 and D2 of the functions. Let R denote the set of all real numbers.

b Determine the average rate of change of the functions over the interval [0,216] and choose the correct statements.

c Graph the functions and describe their end behavior.

### Hint

a The domain of a radical function depends on the index of the root. Does a negative speed make sense?

b Use the formula for the average rate of change of a function.

c Determine the end behavior of functions based on their graphs.

### Solution

a The domains of both functions will be determined one at a time. Start by simplifying the first function rule. To do so, use the fact that μ, the coefficient of friction, is equal to 0.6.

Because the square root of a negative number is imaginary, the input of a square root function must be greater than or equal to 0.

18d≥0⇔d≥0

This means that the domain of the first function is all real numbers greater than or equal to 0. The domain can be written in set-builder notation.

D1={d∣d≥0}

Next, consider the second function. The Product Property of Radicals can be used to simplify its function rule.

v2(p)=838p

v2(p)=163p

The cube root of a negative number is a real number. Mathematically, the domain of this function includes all real numbers R. However, here, a negative root would lead to a negative power and, consequently, a negative speed. These do not make sense for a car. Therefore, the domain is all real numbers greater than or equal to 0.

D2={p∣p≥0}

b The average rate of change of the function f(x) over the interval [x1,x2] can be calculated by using the following formula.

AverageRateofChange=x2−x1f(x2)−f(x1)

To find the average rate of change of v1 and v2 over the interval [0,216], first, the values v1(0), v1(216), v2(0), and v2(216) will be calculated in a table of values. Use a calculator if necessary.

Input | 18d | v1(d) | 163p | v2(p) |
---|---|---|---|---|

0 | 18(0) | 0 | 1630 | 0 |

216 | 18(216) | ≈62.35 | 163216 | 96 |

Now, both average rates of change over the interval [0,216] can be calculated by substituting the obtained values.

Function | x2−x1f(x2)−f(x1) | Substitute and Evaluate |
---|---|---|

v1 | 216−0v1(216)−v1(0) | 216−062.35−0≈0.29 |

v2 | 216−0v2(216)−v2(0) | 216−096−0≈0.44 |

It can be concluded that, on average, the increase in speed of the second model is greater than the increase in speed of the first model. Also, recall that the average rate of change can be thought of as the slope of the corresponding linear function.

On average, if the input of a function increases by **one unit**, the value of the function increases by the **average rate of change**, depending on the interval considered.

This means that the speed of the car increases by about 0.29 for each meter of braking distance, or skid marks left on the ground by braking, and by about 0.44 miles for each unit of horsepower. Therefore, options **A** and **D** are correct.

c Each function will be graphed one at a time by using a table of values. In Part A, it was determined that the domain of v1 is all non-negative real numbers. Substitute such numbers into the function rule.

d | 18d | v1(d) |
---|---|---|

0 | 18(0) | 0 |

2 | 18(2) | 6 |

4 | 18(4) | ≈8.5 |

6 | 18(6) | ≈10.4 |

8 | 18(8) | 12 |

10 | 18(10) | ≈13.4 |

Now, plot the points on a coordinate plane and connect them with a smooth curve.

Based on the graph, the value of the function keeps increasing without bound, so the end behavior of the function is up.

Also, the function is not defined for negative values of d.

Asd→+∞,v1(d)→+∞

Next, just as for the first function, calculate the values of the second function.

p | 163p | v2(p) |
---|---|---|

-8 | 163-8 | -32 |

-6 | 163-6 | ≈-29.1 |

-4 | 163-4 | ≈-25.4 |

-2 | 163-2 | ≈-20.2 |

0 | 1630 | 0 |

2 | 1632 | ≈20.2 |

4 | 1634 | ≈25.4 |

6 | 1636 | ≈29.1 |

8 | 1638 | 32 |

Similarly, plot the obtained points and connect them with a smooth curve.

By analyzing the left and right ends of the graph, the end behavior of the function can be concluded.

Asp→+∞,v2(p)→+∞,Asp→-∞,v2(p)→-∞

To summarize, the correct options are **B**, **D**, and **E**.