# Cal 1

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Verify the given linear approximation at
a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.)

(1 + 2x)^1/4 ≈ 1 + 1/2x

• Cal 1 -

using the binomial expansion,

(1+2x)^(1/4) = 1^(1/4) + (1/4)(1^(-3/4))(2x)^1 + ...
= 1 + 1/2 x + ...

since ∆y/∆x ~= dy/dx, ∆x ~= ∆y/y'

y' = (1/4)(2)(1+2x)^(-3/4) = 1/2 (1+2x)^(-3/4)
y'(0) = 1/2
∆x = 0.1/(1/2) = 0.2

So, -.2 <= x < 0.2

Check:
(1-.4)^(1/4) = 0.880011
1+x/2 = 1-.1 = 0.9
error = 0.02

(1+.4)^(1/4) = 1.0877
1+x/2 = 1+.4/2 = 1.2
error = 0.12

Hmmm. Looks like I was off a bit, on the + side, and way too strict on the - side.

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