Compare the estimated average rate of change of the exponential function y=9(1/3)^x

and the quadratic function y=9x^2+1/3x
. Which function has a negative estimated average rate of change over the interval [0.1,0.6]
?(1 point)
Responses

neither function
neither function

the exponential function
the exponential function

both functions
both functions

the quadratic function

1 answer

To find the estimated average rate of change of both functions over the interval \([0.1, 0.6]\), we will calculate the average rate of change using the formula:

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

where \(a = 0.1\) and \(b = 0.6\).

For the exponential function \(y = 9(1/3)^x\):

  1. Calculate \(f(0.1)\): \[ f(0.1) = 9\left(\frac{1}{3}\right)^{0.1} \approx 9 \times 0.933 = 8.397 \]

  2. Calculate \(f(0.6)\): \[ f(0.6) = 9\left(\frac{1}{3}\right)^{0.6} \approx 9 \times 0.481 = 4.329 \]

  3. Compute the average rate of change: \[ \text{Average Rate of Change} = \frac{4.329 - 8.397}{0.6 - 0.1} = \frac{-4.068}{0.5} = -8.136 \]

For the quadratic function \(y = 9x^2 + \frac{1}{3}x\):

  1. Calculate \(f(0.1)\): \[ f(0.1) = 9(0.1)^2 + \frac{1}{3}(0.1) = 9(0.01) + \frac{0.1}{3} = 0.09 + 0.0333 \approx 0.1233 \]

  2. Calculate \(f(0.6)\): \[ f(0.6) = 9(0.6)^2 + \frac{1}{3}(0.6) = 9(0.36) + 0.2 = 3.24 + 0.2 = 3.44 \]

  3. Compute the average rate of change: \[ \text{Average Rate of Change} = \frac{3.44 - 0.1233}{0.6 - 0.1} = \frac{3.3167}{0.5} \approx 6.6334 \]

Conclusion:

  • The average rate of change for the exponential function is approximately \(-8.136\) (negative).
  • The average rate of change for the quadratic function is approximately \(6.6334\) (positive).

Thus, the function with a negative estimated average rate of change over the interval \([0.1, 0.6]\) is:

the exponential function.