Question

question 14 (1 point)

for (f(x) = -x^2 + 1) which value has the largest magnitude?

average rate of change from (x = -4) to (x = 0)

instantaneous rate of change at (x = 3)

instantaneous rate of change at (x = 0)

average rate of change from (x = -1) to (x = 1)

Explanation:

Calculate average rate of change from x = -4 to x = 0

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

$$\text{Average Rate} = \frac{f(b) - f(a)}{b - a}$$

For \(f(x) = -x^2 + 1\) on \([-4, 0]\):

$$f(0) = -(0)^2 + 1 = 1$$

$$f(-4) = -(-4)^2 + 1 = -16 + 1 = -15$$

$$\text{Average Rate} = \frac{1 - (-15)}{0 - (-4)} = \frac{16}{4} = 4$$

The magnitude of this rate is \(|4| = 4\).

Calculate instantaneous rate of change at x = 3

Using the Instantaneous Rate of Change knowledge point, we find the derivative \(f'(x)\) of \(f(x) = -x^2 + 1\):

$$f'(x) = -2x$$

Evaluating at \(x = 3\):

$$f'(3) = -2(3) = -6$$

The magnitude of this rate is \(|-6| = 6\).

Calculate instantaneous rate of change at x = 0

Using the Instantaneous Rate of Change knowledge point, we evaluate the derivative at \(x = 0\):

$$f'(0) = -2(0) = 0$$

The magnitude of this rate is \(|0| = 0\).

Calculate average rate of change from x = -1 to x = 1

For \(f(x) = -x^2 + 1\) on \([-1, 1]\):

$$f(1) = -(1)^2 + 1 = 0$$

$$f(-1) = -(-1)^2 + 1 = 0$$

$$\text{Average Rate} = \frac{0 - 0}{1 - (-1)} = 0$$

The magnitude of this rate is \(|0| = 0\).

Compare the magnitudes of the rates of change

We compare the absolute values of each calculated rate of change:

• Option 1: Magnitude is \(4\)
• Option 2: Magnitude is \(6\)
• Option 3: Magnitude is \(0\)
• Option 4: Magnitude is \(0\)

The largest magnitude is \(6\), which corresponds to the instantaneous rate of change at \(x = 3\).

Answer:

• average rate of change from x = -4 to x = 0
• instantaneous rate of change at x = 3 (Correct answer)
• instantaneous rate of change at x = 0
• average rate of change from x = -1 to x = 1