Question

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%

Answer 1

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

Answer 2

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 <b>y ≂ 18.778x - 7.389</b>

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.

Answer 3

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