#### Answer

(a) $g(1.95) \approx -4.15$
$g(2.05) \approx -3.85$
(b) The approximations found in part (a) are too small.

#### Work Step by Step

(a) When $x = 2$:
$g(2) = -4$
$g'(2) = \sqrt{x^2+5} = \sqrt{2^2+5} = 3$
We can find the linear approximation at $a=2$:
$g(x) \approx g(a)+g'(a)(x-a)$
$g(x) \approx g(2)+g'(2)(x-2)$
$g(x) \approx (-4)+3(x-2)$
$g(x) \approx 3x-10$
We can approximate $g(1.95)$:
$g(x) \approx 3x-10$
$g(1.95) \approx 3(1.95)-10$
$g(1.95) \approx -4.15$
We can approximate $g(2.05)$:
$g(x) \approx 3x-10$
$g(2.05) \approx 3(2.05)-10$
$g(2.05) \approx -3.85$
(b) As $x$ increases, the value of $g'(x)$ increases.
Therefore, $g(x)$ is concave up.
If a line is tangent to a function that is concave up, then the tangent line is below the function, so the approximations found in part (a) are too small.