Answer
a) 3.6
b) -0.777778
Work Step by Step
a) $f(x) = x^{2}-1$
$f'(x) = 2x$
$f'(x_{0}) = 2(x_{0})$
As $x_{0} = 1.8$
$f(1.8) = 2(1.8)$
$f(1.8) = 3.6$
b) $f(x) = \frac{x^{2}}{x-2}$
$f'(x) = \frac{(x-2)(2x)-x^{2}(1)}{(x-2)^{2}}$
After simplification:
$f'(x) = \frac{x^{2}-4x}{(x-2)^{2}}$
$f'(x_{0}) = \frac{(x_{0})^{2}-4x_{0}}{(x_{0}-2)^{2}}$
As $x_{0} = 3.5$
$f'(3.5) = \frac{(3.5)^{2}-4(3.5)}{(3.5-2)^{2}}$
$f(3.5) = -\frac{7}{9} \approx -0.777778$
