1.

Find the estimated average rates of change of the exponential function pictured above and the quadratic function f(x)=x2−20
over the interval [-9,2]. Show the work for both functions.

Graphed function:

f(x)
:



2.

Time (hours)

0

4

7

9

12

Distance (miles)

0

248

449

565

757

The Diaz family went on a road trip for vacation. The table includes a cumulative chart of the time they traveled compared to the distance they drove. Over which interval was their average speed the slowest and fastest: [0, 4], [4, 7], [7, 9], or [9, 12]? Show work for each interval's average speed.

Slowest interval: [
,
]

Fastest interval: [
,
]

1 answer

To tackle both of your queries, let's start with the first part regarding the average rates of change of the functions.

1. Average Rates of Change

a. For the quadratic function \( f(x) = x^2 - 20 \):

The average rate of change of a function \( f \) over the interval \([a, b]\) is given by:

\[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \]

For our quadratic function and the interval \([-9, 2]\):

  • \( a = -9 \)
  • \( b = 2 \)

First, we need to calculate \( f(-9) \) and \( f(2) \):

\[ f(-9) = (-9)^2 - 20 = 81 - 20 = 61 \]

\[ f(2) = (2)^2 - 20 = 4 - 20 = -16 \]

Now, apply these values to the average rate of change formula:

\[ \text{Average Rate of Change} = \frac{f(2) - f(-9)}{2 - (-9)} = \frac{-16 - 61}{2 + 9} = \frac{-77}{11} = -7 \]

So, the estimated average rate of change for the quadratic function \( f(x) = x^2 - 20 \) over the interval \([-9, 2]\) is -7.

b. For the exponential function (assuming it resembles a typical \( f(x) = a^x \) function):

Since you mentioned an "exponential function pictured above," but we don't have any specific equation or values for this, let's assume it is represented as \( f(x) = e^x \) (you can substitute this with the actual function if different).

We similarly need to find \( f(-9) \) and \( f(2) \):

\[ f(-9) = e^{-9} \quad (\text{Approximates to } 0.000123) \] \[ f(2) = e^2 \quad (\text{Approximates to } 7.389) \]

Now, we can calculate the average rate of change:

\[ \text{Average Rate of Change} = \frac{f(2) - f(-9)}{2 - (-9)} \]

Calculating it gives:

\[ \text{Average Rate of Change} = \frac{7.389 - 0.000123}{2 + 9} = \frac{7.388877}{11} \approx 0.671 \]

So, the estimated average rate of change for the exponential function over the interval \([-9, 2]\) is approximately 0.671.

2. Average Speed Intervals

To find the average speed for each interval, the average speed can be calculated using the following formula:

\[ \text{Average Speed} = \frac{\text{Distance traveled}}{\text{Time taken}} \]

We need to calculate the average speed for the intervals \([0, 4]\), \([4, 7]\), \([7, 9]\), and \([9, 12]\):

  • Interval \([0, 4]\):

    • Distance traveled: \( 248 - 0 = 248 \) miles
    • Time taken: \( 4 - 0 = 4 \) hours

    \[ \text{Average Speed} = \frac{248}{4} = 62 \text{ miles/hour} \]

  • Interval \([4, 7]\):

    • Distance traveled: \( 449 - 248 = 201 \) miles
    • Time taken: \( 7 - 4 = 3 \) hours

    \[ \text{Average Speed} = \frac{201}{3} = 67 \text{ miles/hour} \]

  • Interval \([7, 9]\):

    • Distance traveled: \( 565 - 449 = 116 \) miles
    • Time taken: \( 9 - 7 = 2 \) hours

    \[ \text{Average Speed} = \frac{116}{2} = 58 \text{ miles/hour} \]

  • Interval \([9, 12]\):

    • Distance traveled: \( 757 - 565 = 192 \) miles
    • Time taken: \( 12 - 9 = 3 \) hours

    \[ \text{Average Speed} = \frac{192}{3} = 64 \text{ miles/hour} \]

Summary of Average Speeds

  • Interval \([0, 4]\): 62 mph
  • Interval \([4, 7]\): 67 mph
  • Interval \([7, 9]\): 58 mph
  • Interval \([9, 12]\): 64 mph

Results for Fastest and Slowest Intervals:

  • Slowest interval: \([7, 9]\) at 58 mph
  • Fastest interval: \([4, 7]\) at 67 mph

Thus, we conclude:

  • Slowest interval: \([7, 9]\)
  • Fastest interval: \([4, 7]\)