Any curve can be approximated by a straight line, in a small enough interval. So, we want the tangent line at x=0, which will be very close to the curve, if we stay close enough to x=0.
y=(1+x)^(1/4)
y(0) = 1
y' = 1/4 * (1+x)^(-3/4)
= 1/[4(1+x)^3]
y'(0) = 1/4
So, the line y = x/4 + 1 is tangent to f(x) at x=0
let x = -0.05
y = -.0125 + 1 = 0.9875
If x = .1
y = .025 + 1 = 1.025
Just to check,
.95^(1/4) = 0.9873
1.1^(1/4) = 1.0241
Since |.95 - 1| < |1.1 - 1| the approximation is better
Find a linear approximation of the function f(x)=(1+x)^(1/4) at a=1, and use it to approximate the numbers (.95)^(1/4) and (1.1)^(1/4).
Round your answers to the nearest thousandth
Cheers in advance!
1 answer
1 answer