Answer
a. $1-6x$,
b. $2+2x$,
c. $1-\frac{1}{2}x$,
d. $\sqrt 2 (1+\frac{1}{4}x^2)$,
e. $\sqrt[3] 4(1+\frac{1}{4}x)$,
f. $1-\frac{2x}{6+3x}$
Work Step by Step
Use the linear approximation is:
$(1+x)^k=1+kx$
a. $f(x)=(1-x)^6=1+6(-x)=1-6x$,
b. $f(x)=\frac{2}{1-x}=2(1-x)^{-1}=2(1-1(-x))=2+2x$,
c. $f(x)=\frac{1}{\sqrt {1+x}}=(1+x)^{-1/2}=1-\frac{1}{2}x$,
d. $f(x)=\sqrt {2+x^2}=\sqrt 2\sqrt {1+x^2/2}=\sqrt 2 (1+x^2/2)^{1/2}=\sqrt 2 (1+\frac{1}{2}(x^2/2))=\sqrt 2 (1+\frac{1}{4}x^2)$,
e. $f(x)=(4+3x)^{1/3}=4^{1/3}(1+3x/4)^{1/3}=4^{1/3}(1+\frac{1}{3}(3x/4))=4^{1/3}(1+\frac{1}{4}x)$,
f. $f(x)=\sqrt[3] {(1-\frac{x}{2+x})^2}=(1-\frac{x}{2+x})^{2/3}=1+\frac{2}{3}(\frac{-x}{2+x})=1-\frac{2x}{6+3x}$
