Question

find the range of f(x)= (x^2 + 1)/(x^2 + 5) if the domain is (-1/2<= x<= 2)

Answer 1

Oh, that fellow is VERY positive. Can you even get it down to zero?

Answer 2

Find f'(x) and the critical point(s) in the domain [-1/2, 2].
Find where the function is increasing/decreasing.

Answer 3

\[
f(x)= \frac{(x^2 + 1)}{(x^2 + 5)} \\
f'(x) = \frac{(x^2+5)(2x)-(x^2+1)(2x)}{(x^2+5)^2} =
\frac{2x(x^2+5-x^2-1)}{(x^2+5)^2} = \frac{8x}{(x^2+5)^2} = 0\\
x = 0 \text{ is a critical point.} \\
x \lt 0, f'(x) < 0, f(x) \text{ is decreasing.} \\
x \gt 0, f'(x) > 0, f(x) \text{ is increasing.} \\
\]

Answer 4

f(x) attains minimum when x = 0 and f(0) = (0+1)/(0+5) = 1/5 = 0.2
In [-1/2, 0), f(x) is decreasing. So maximum at x = -1/2.
f(-1/2) = (1/4+1)/(1/4+5) = 5/21.

In (0, 2], f(x) is increasing. So maximum at x = 2.
f(2) = (4+1)/(4+5) = 5/9.

5/9 > 5/21

So the range of f(x) in the domain [-1/2, 2] is [1/5, 5/9].

Answer 5

The question is still open so I assume you have probably not covered derivatives, critical points, etc. yet. You can also find the range algebraically:
\[
f(x)= \frac{(x^2 + 1)}{(x^2 + 5)} = \frac{(x^2 + 5-4)}{(x^2 + 5)} =
\frac{(x^2 + 5)}{(x^2 + 5)} - \frac{(4)}{(x^2 + 5)} =
1 - \frac{4}{(x^2 + 5)} \\
\]The minimum value of \((x^2+5)\) is when x = 0 and it is 5. So the maximum value of \(\large \frac{4}{(x^2+5)}\) is 4/5 and the minimum value of f(x) is 1 - 4/5 = 1/5.
Like before, f(-1/2) = 5/21; f(2) = 5/9. 5/9 > 5/21.

So the range of f(x) in the domain [-1/2, 2] is [1/5, 5/9].