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
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%
3 answers
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.
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
I messed up in my 2nd last line
should have been:
f(1.1) = 4(1.1) + e^(2.2) = appr 13.425
1 answer