Question

(1 point) Use the linear approximation to estimate (1.02)³(-3.02)³ ≈ Compare with the value given by a calculator and compute the percentage error: Error = %

Answer 1

To estimate the value of (1.02)³(-3.02)³ using linear approximation, find the linear approximation of each term separately. Multiply the approximations to get the estimated value. Compare this value with the actual value from a calculator and calculate the percentage error.

Step-by-step explanation:

To estimate the value of (1.02)³(-3.02)³ using linear approximation, we need to find the linear approximation of each term separately. The linear approximation of (1.02)³ is 1 + 3(1.02 - 1) = 1 + 3(0.02) = 1.06, and the linear approximation of (-3.02)³ is -3 + 3(-3.02 + 3) = -3 + 3(-0.02) = -3 - 0.06 = -3.06. Multiplying these two approximations gives us an estimated value of 1.06 * (-3.06) = -3.2436.

To compare this with the value given by a calculator, we can calculate (1.02)³(-3.02)³ directly using a calculator, which gives us an actual value of -3.417456.

To compute the percentage error, we can use the formula: Error = ((Estimated Value - Actual Value) / Actual Value) * 100%. Substituting the values, we get ((-3.2436 - (-3.417456)) / (-3.417456)) * 100%. Simplifying this gives us an error of 5.0881% (rounded to four decimal places).

Answer 2

To estimate the product of (1.02)^3(-3.02)^3, we use linear approximation for small changes around 1 and -3 and find the estimated value is -29.2324. We then compare this with the actual calculator value and use the formula to calculate the percentage error.

Step-by-step explanation:

To estimate the value of (1.02)^3(-3.02)^3 using a linear approximation, we can apply the formula for the derivative at a point, which gives us the best linear approximation near that point. For small changes in the input, we can assume the function behaves similarly to its tangent line. We will use the fact that for any function f(x) that is differentiable at x = a, we can say that f(x) ≈ f(a) + f'(a)(x-a).

The function we are considering here is f(x) = x^3. In this case, we're looking at points near 1 for the first part (1.02)^3 and -3 for the second part (-3.02)^3. The derivative of f(x) = x^3 is f'(x) = 3x^2. Therefore, for a small change around 1 and -3, we can use the approximations:

• (1.02)^3 ≈ 1^3 + 3(1^2)(1.02-1) = 1 + 3(0.02) = 1.06
• (-3.02)^3 ≈ (-3)^3 + 3(-3)^2(-3.02 + 3) = -27 + 3(9)(-0.02) = -27 - 0.54 = -27.54

Now, by multiplying these two approximations, we get an estimate:

(1.02)^3(-3.02)^3 ≈ 1.06 * (-27.54) ≈ -29.2324

To find the percentage error, we compare this with the value given by a calculator:

Actual value from calculator: (1.02)^3(-3.02)^3

Estimated value from linear approximation: -29.2324

Error = (Actual value - Estimated value) / Actual value * 100%