Answer
(a) $f(x) \approx \frac{25}{4}-\frac{3}{4}~x$
(b) We can see a sketch of the graph below.
(c) The linear approximation is accurate to within $0.1$ for the values $2.3 \leq x \leq 3.6$
Work Step by Step
(a) $f(x) = \sqrt{25-x^2}$
$f'(x) = \frac{1}{2}(25-x^2)^{-1/2}~(-2x)$
$f'(x) = -\frac{x}{\sqrt{25-x^2}}$
When $x = 3$:
$f(3) = \sqrt{25-3^2} = 4$
$f'(3) = -\frac{3}{\sqrt{25-3^2}} = -\frac{3}{4}$
We can find the linear approximation near $a=3$:
$f(x) \approx f(a)+f'(x-a)$
$f(x) \approx f(3)+f'(x-3)$
$f(x) \approx 4-\frac{3}{4}(x-3)$
$f(x) \approx \frac{25}{4}-\frac{3}{4}~x$
(b) We can see a sketch of the graph below.
(c) With the help of the graph, we can calculate that the linear approximation is accurate to within $0.1$ for the values $2.3 \leq x \leq 3.6$
We can verify this:
When $x = 2.3$:
$\frac{25}{4}-\frac{3}{4}~(2.3) = 4.53$
$f(2.3) = \sqrt{25-2.3^2} = 4.44$
When $x = 3.6$:
$\frac{25}{4}-\frac{3}{4}~(3.6) = 3.55$
$f(3.6) = \sqrt{25-3.6^2} = 3.47$
