If the local linear approximation of f(x) = 2cos x + e2x at x = 2 is used to find the approximation for f(2.1), then the % error of this approximation is...
greater than 15%
between 11% and 15%
between 5% and 10%
between 0% and 4%
To find the approximation for f(2.1) using the local linear approximation, we will first need to find the equation of the tangent line to f(x) at x = 2.
To do this, we'll find the first derivative of f(x), which is:
f'(x) = -2sin(x) + 2e^(2x)
Then, we can find the slope of the tangent line by evaluating f'(x) at x = 2:
f'(2) = -2sin(2) + 2e^(2*2) ≈ -2(0.909) + 2(54.598) ≈ -1.818 + 109.196 ≈ 107.378
So, the equation of the tangent line at x = 2 is given by:
y = f(2) + f'(2)(x - 2)
Now, let's evaluate f(2) and substitute the values into the equation to find the equation of the tangent line.
f(2) = 2cos(2) + e^(2*2) ≈ 2(-0.416) + 16.389 ≈ -0.832 + 16.389 ≈ 15.557
Therefore, the equation of the tangent line is:
y = 15.557 + 107.378(x - 2)
Now, we can use this equation to approximate f(2.1). We substitute x = 2.1 into the equation:
f(2.1) ≈ 15.557 + 107.378(2.1 - 2)
≈ 15.557 + 107.378(0.1)
≈ 15.557 + 10.738
≈ 26.295
Next, we need to find the percent error between the approximation f(2.1) ≈ 26.295 and the actual value of f(2.1).
To do that, we'll need to find the actual value of f(2.1) using the original function f(x) = 2cos(x) + e^(2x):
f(2.1) = 2cos(2.1) + e^(2*2.1) ≈ 2(-0.504) + 47.131 ≈ -1.008 + 47.131 ≈ 46.123
The percent error is calculated as:
Percent error = | (Approximate value - Actual value) / Actual value | * 100
Percent error = |(26.295 - 46.123) / 46.123| * 100
≈ |-19.828 / 46.123| * 100
≈ 0.430 * 100
≈ 4.30%
Therefore, the % error of this approximation is between 0% and 4%.
To find the percent error of the approximation using the local linear approximation, we can compare the actual value of f(2.1) with the approximate value obtained from the linear approximation.
First, let's find the actual value of f(2.1) using the function f(x) = 2cos x + e^2x.
f(2.1) = 2cos(2.1) + e^(2*2.1)
Using a calculator, we can find that cos(2.1) is approximately -0.5048 and e^(2*2.1) is approximately 109.86.
Therefore, f(2.1) ≈ 2*(-0.5048) + 109.86 ≈ 108.85
Next, let's find the approximate value of f(2.1) using the local linear approximation at x = 2.
The local linear approximation of f(x) at x = 2 is given by the equation:
L(x) = f(a) + f'(a)(x-a)
Here, a = 2 and f(a) = f(2) = 2cos(2) + e^(2*2).
Using a calculator, we find cos(2) is approximately -0.4161 and e^(2*2) is approximately 54.5981.
Therefore, f(2) ≈ 2*(-0.4161) + 54.5981 ≈ 54.766
To find f'(a), we differentiate the function f(x) = 2cos x + e^(2x) with respect to x:
f'(x) = -2sin x + 2e^(2x)
Evaluating f'(2), we find:
f'(2) = -2sin(2) + 2e^(2*2)
Using a calculator, we find sin(2) is approximately 0.9093 and e^(2*2) is approximately 54.5981.
Therefore, f'(2) ≈ -2*(0.9093) + 2*(54.5981) ≈ 108.761
Now, let's substitute these values into the local linear approximation equation to find the approximate value of f(2.1):
L(2.1) = f(2) + f'(2)(2.1-2)
L(2.1) ≈ 54.766 + 108.761*(2.1-2) ≈ 54.766 + 108.761*0.1 ≈ 54.766 + 10.8761 ≈ 65.6421
Finally, let's calculate the percent error:
Percent error = (|Approximate value - Actual value| / |Actual value|) * 100
Percent error = (|65.6421 - 108.85| / |108.85|) * 100
Percent error ≈ (|-43.2079| / 108.85) * 100 ≈ (43.2079 / 108.85) * 100 ≈ 39.66%
Since the percent error is approximately 39.66%, which is greater than 15%, the answer is: greater than 15%