AP Calculus AB Exam Question:

Consider the function f(x) = e^x - ln(x^2), where e is the base of the natural logarithm.

(a) Using linear approximation, find an approximation for f(1.1) and determine the percent error in the approximation when compared to the actual value.

(b) Use differentials to estimate the change in f if x undergoes a small change from x = 1 to x = 1.05.

Solution:

(a) To find an approximation for f(1.1) using linear approximation, we can start by using the equation for linear approximation:

L(x) = f(a) + f'(a)(x-a)

Where f(a) represents the value of the function at point a, f'(a) is the derivative of the function evaluated at point a, and (x-a) is the difference between the desired point x and the point a.

To find f(1.1), we will use a = 1 as the starting point for linear approximation.

First, let's find f'(x):

f(x) = e^x - ln(x^2)

Taking the derivative with respect to x:

f'(x) = e^x - 2/x

Now, evaluate f'(1):

f'(1) = e^1 - 2/1 = e - 2

Next, substitute the values into the linear approximation equation:

L(x) = f(a) + f'(a)(x-a)

L(1.1) = f(1) + f'(1)(1.1 - 1) = f(1) + (e - 2)(0.1)

Now, substitute a = 1 into f(x):

f(1) = e^1 - ln(1^2) = e - 0 = e

Finally, substitute the values back into the linear approximation equation:

L(1.1) = e + (e - 2)(0.1)

Calculating the value, we get:

L(1.1) = e + (e - 2)(0.1) ≈ 0.1e + 0.2

The actual value of f(1.1) can be found by substituting x = 1.1 into the function:

f(1.1) = e^(1.1) - ln((1.1)^2)

Using a calculator, we find:

f(1.1) ≈ 2.5742

To calculate the percent error, we use the formula:

Percent error = |(actual value - approximate value) / actual value| * 100

Percent error = |(2.5742 - (0.1e + 0.2)) / 2.5742| * 100

Calculating the value, we get:

Percent error ≈ |0.1258 / 2.5742| * 100

Percent error ≈ 4.891%

Therefore, the approximate value of f(1.1) using linear approximation is 0.1e + 0.2, and the percent error compared to the actual value is approximately 4.891%.

(b) To estimate the change in f if x undergoes a small change from x = 1 to x = 1.05 using differentials, we can use the formula:

df ≈ f'(a) * dx,

where dx is the small change in x.

To find df, we need to find f'(a) for a = 1:

f'(x) = e^x - 2/x

f'(1) = e^1 - 2/1 = e - 2

Now, substitute the values into the differentials formula:

df ≈ (e - 2) * dx.

Given that x changes from 1 to 1.05 (dx = 1.05 - 1 = 0.05), we can calculate the estimated change in f:

df ≈ (e - 2) * 0.05.

Calculating the value, we get:

df ≈ 0.05e - 0.1.

Therefore, the estimated change in f when x undergoes a small change from x = 1 to x = 1.05 using differentials is approximately 0.05e - 0.1.