I need help here please thanks

asked Feb 10 in Calculus Answers by Joseph

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9. Several ways to solve these.

x³=8±0.5=8(1±0.5/8)=8(1±1/16)

Taking cube roots of each side:

x=∛(8(1±1/16))=2(1±1/16)^(1/3)=2(1±1/48) as first degree approximation of binomial expansion;

or x=2∛(1±0.0625)=(2∛0.9375,2∛1.0625)=2(0.97872×2,1.02041×2)=(1.9574,2.0408) approx.

The binomial method makes x=2±1/24=2±0.042 approx, so x=(1.96,2.04).

Another way is to calculate ∛7.5=1.9574 and ∛8.5=2.0408 so x=(1.9574,2.0408) approx.

The binomial method will be more accurate for a margin of 0.05 compared to 0.5:

x=2(1±0.05/8)^(1/3)=2(1±0.05/24)=2(1±0.00208)=(1.99583,2.00417).

which compares well with (∛7.95,∛8.05)=(1.99582,2.00416).

10 The general case is √(x±h)=4±k where h is the variation in x corresponding to k, the variation in 4.

Squaring: x±h=16±8k+k² and when x=16, ±h=±8k+k².

When k=1, h=-7 or 9 and x=(16-7,16+9)=(9,25). Variation in x is these values of h.

When k=0.1, h=-0.79 or 0.81 and x=(16-0.79,16+0.81)=(15.21,16.81)=(3.9²,4.1²). Variation in x is these values in h.

answered Feb 10 by Rod Top Rated User (538,080 points)

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