If the local linear approximation of f(x)=4x+e^2x at x=1 is used to find the approximation for f(1.1), then the % error of this approximation is
Between 0% and 4%
Between 5% and 10%
Between 11% and 15%
Greater than 15%
To find the local linear approximation of f(x) = 4x + e^(2x) at x = 1, we first need to determine the equation of the tangent line to the graph of f(x) at x = 1.
The slope of the tangent line can be found by taking the derivative of f(x) with respect to x and evaluating it at x = 1.
f'(x) = 4 + 2e^(2x)
Evaluate f'(x) at x = 1:
f'(1) = 4 + 2e^(2*1) = 4 + 2e^2
So, the slope of the tangent line at x = 1 is 4 + 2e^2.
To find the approximation for f(1.1) using the local linear approximation, we use the point-slope form of a linear equation:
y - y1 = m(x - x1)
where (x1, y1) is the point (1, f(1)) and m is the slope of the tangent line at x = 1.
Using the point (1, f(1)), which is (1, 4e^2 + e^2), we get:
y - (4e^2 + e^2) = (4 + 2e^2)(x - 1)
Now, we can substitute x = 1.1 into the equation above to find the approximation for f(1.1):
f(1.1) - (4e^2 + e^2) = (4 + 2e^2)(1.1 - 1)
Simplifying this equation gives:
f(1.1) = (4 + 2e^2)(0.1) + 4e^2 + e^2
To calculate the percentage error of this approximation, we can compare it to the actual value of f(1.1).
To find the actual value of f(1.1), we substitute x = 1.1 into the original function:
f(1.1) = 4(1.1) + e^(2*1.1)
Now, we can calculate the percentage error using the formula:
% error = |(approximation - actual value) / actual value| * 100
Substituting the expressions for the approximation and actual value of f(1.1) into the formula above and simplifying, we get:
% error = |((4 + 2e^2)(0.1) + 4e^2 + e^2 - (4(1.1) + e^(2*1.1))) / (4(1.1) + e^(2*1.1))| * 100
We can calculate the value of this expression to find the percentage error.
df/dx = f'(x) = 4+ 2x e^2x
f(1) = 4 + e^2 = 4 + 7.389 = 11.389
f'(1) = 4 + 2 e^2 = 4 + 14.778 = 18.778
f(1.1) = f(1) + 0.1 f'(1)
= 11.389 + 1.878 = 13.27
real = 4.4 + 9.02 = 13.42
f(x)=4x+e^2x
f ' (x) = 4 + 2e^(2x)
f ' (1) = 4 + 2e^2
f(1) = 4 + e^2
linear equation: y - (4 + e^2) = (4+2e^2)(x-1)
y - 4 - e^2 = 4x + (2e^2)x - 4 - 2e^2
y = 4x + 2e^2 x - e^2 OR y ≂ 18.778x - 7.389
when x = 1.1, y ≂ 18.778(1.1) - 7.389 = appr 13.267
from original function:
f(1.1) = 4(1.1) + 2e^(2.2) = appr 13.65
see what you can do with that.
go with Damon's
I messed up in my 2nd last line
should have been:
f(1.1) = 4(1.1) + e^(2.2) = appr 13.425