1. ## linear approximation

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

$\displaystyle \sqrt{1+x}\approx1+\frac{1}{2}x$

which assuming means
$\displaystyle \sqrt{1+x}-0.1<1+\frac{1}{2}x<\sqrt{1+x}+0.1$

but couldn't get the given answer of
$\displaystyle -0.69<x<1.09$

2. First, we notice that the domain is $\displaystyle x \ge -1$.

$\displaystyle 1+\frac{x}{2}<\sqrt{1+x}+0.1$

$\displaystyle \implies~ 0.9+0.5x<\sqrt{1+x}$

Since both sides of the inequality are always non-negative over the domain $\displaystyle x \ge -1$:

$\displaystyle \implies~ (0.9+0.5x)^2<1+x$

$\displaystyle \implies~ 0.81+0.9x+0.25x^2<1+x$

$\displaystyle \implies~ 0.25x^2 - 0.1x - 0.19<0$

Solving this gives $\displaystyle -0.694427<x<1.09443$ which matches the solution you gave.

By the way, the other side doesn't actually give any restriction:

$\displaystyle \sqrt{1+x}-0.1<1+\frac{x}{2}$

$\displaystyle \implies~ \sqrt{1+x}<1.1+0.5x$

$\displaystyle \implies~ 1+x < (1.1+0.5x)^2$

$\displaystyle \implies~ 1+x < 1.21 + 1.1x + 0.25x^2$

$\displaystyle \implies~ 0 < 0.25x^2 + 0.1x + 0.21$

Which is always true, so there is no additional restriction here.