For |x|

Question

(C) Given that $|x| < \frac{4}{3}$.We have $\frac{1}{(4-3 x)^{\frac{1}{2}}} = (4-3 x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \left(1-\frac{3 x}{4}\right)^{

Vedclass · Accepted Answer

$\frac{1}{2}+\frac{3 x}{16}+\frac{27 x^2}{256}$

Vedclass · Answer

(C) Given that $|x| We have $\frac{1}{(4-3 x)^{\frac{1}{2}}} = (4-3 x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}}$.Using the binomial expansion $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!} u^2 + \dots$, where $u = -\frac{3x}{4}$ and $n = -\frac{1}{2}$:$\frac{1}{2} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2} \left[ 1 + \left(-\frac{1}{2}\right) \left(-\frac{3x}{4}\right) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2} \left(-\frac{3x}{4}\right)^2 + \dots \right]$$= \frac{1}{2} \left[ 1 + \frac{3x}{8} + \frac{3}{8} \cdot \frac{9x^2}{16} + \dots \right]$$= \frac{1}{2} \left[ 1 + \frac{3x}{8} + \frac{27x^2}{128} + \dots \right]$$= \frac{1}{2} + \frac{3x}{16} + \frac{27x^2}{256} + \dots$

For $|x| < \frac{4}{3}$, the approximate value of $\frac{1}{(4-3 x)^{\frac{1}{2}}}$ is

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