Question

In Exercises 9 and \(10,\) use the linear approximation \((1+x)^{k} \approx 1+k x\) to find an approximation for the function \(f(x)\) for values of \(x\) near zero. (a) \(f(x)=(1-x)^{6}\) (b) \(f(x)=\frac{2}{1-x}\) (c) \(f(x)=\frac{1}{\sqrt{1+x}}\)

Short answer

The approximations for the given functions near zero are as follows: (a) \(1-6x\), (b) \(2-2x\), (c) \(1-\frac{1}{2}x\).

Step-by-step solution

Approximate Function (a)

For function (a) \(f(x)=(1-x)^{6}\), we rewrite it to match the base form. Notice it can be written as \((1+(-x))^{6}\). Thus with the approximation \(1+kx\), k is both the exponent 6 and the value of -x, leading to the approximation \(1+6(-x)\) which simplifies to \(1-6x\).

Approximate Function (b)

For function (b) \(f(x)=\frac{2}{1-x}\), we rewrite it as \(2*(1-x)^{-1}\) which is in the form \((1+(-x))^{-1}\) .Thus, k=-1 and 'x' in the approximation formula is -x. Using the approximation \(1+kx\), the approximation of the function becomes \(1-(-1)(-x)\) which simplifies to \(1-x\). We still have the multiplication with 2 ahead, so the final approximation is \(2*(1-x)\) which simplifies to \(2-2x\).

Approximate Function (c)

For function (c) \(f(x)=\frac{1}{\sqrt{1+x}}\), we rewrite it as \((1+x)^{\-1/2}\) where k=-1/2. Here, our 'x' in this problem comes directly from the function (1+x). Thus, the approximation is \(1-kx\) which becomes \(1-\frac{1}{2}x\).