Questions: Consider the curve f(x)=4 x^2, 0 ≤ x ≤ 3. For which of the given intervals is the slope of the secant line equal to 16? [1.9,2.1] [2,2.1] [1.9,2] [0,2]

Solution Steps

To find the interval where the slope of the secant line is equal to 16, we need to calculate the slope of the secant line for each given interval. The slope of the secant line between two points \((x_1, f(x_1))\) and \((x_2, f(x_2))\) on the curve \(f(x)\) is given by:

\[ \text{slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]

We will compute this slope for each interval and check which one equals 16.

We are given the function \( f(x) = 4x^2 \) and the intervals \([1.9, 2.1]\), \([2, 2.1]\), \([1.9, 2]\), and \([0, 2]\). We need to find the interval where the slope of the secant line is equal to 16.

The slope of the secant line between two points \((x_1, f(x_1))\) and \((x_2, f(x_2))\) on the curve \( f(x) \) is given by: \[ \text{slope} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]

We will compute the slope for each interval:

1. For \([1.9, 2.1]\): \[ \text{slope} = \frac{f(2.1) - f(1.9)}{2.1 - 1.9} = \frac{4(2.1)^2 - 4(1.9)^2}{2.1 - 1.9} = \frac{17.64 - 14.44}{0.2} = \frac{3.20}{0.2} = 16 \]

2. For \([2, 2.1]\): \[ \text{slope} = \frac{f(2.1) - f(2)}{2.1 - 2} = \frac{4(2.1)^2 - 4(2)^2}{2.1 - 2} = \frac{17.64 - 16}{0.1} = \frac{1.64}{0.1} = 16.4 \]

3. For \([1.9, 2]\): \[ \text{slope} = \frac{f(2) - f(1.9)}{2 - 1.9} = \frac{4(2)^2 - 4(1.9)^2}{2 - 1.9} = \frac{16 - 14.44}{0.1} = \frac{1.56}{0.1} = 15.6 \]

4. For \([0, 2]\): \[ \text{slope} = \frac{f(2) - f(0)}{2 - 0} = \frac{4(2)^2 - 4(0)^2}{2 - 0} = \frac{16 - 0}{2} = \frac{16}{2} = 8 \]

Final Answer