If f(x) = {(3^(-1), when -1) ≤ x < 0), (tan x/2, when 0 < x < π,), (x/x^2-2, when π ≤ x ≤ 6)

Question

If

\( f(x) = \begin{cases} 3^{-x} -1, & \quad \text{when -1 } \leq x < 0\\ tan(\frac{x}{2},) & \quad \text{when} 0<x< \pi, \\ \frac{x}{x^2-2}, & \quad \text{when}\pi \leq x \leq 6\\ \end{cases} \)

then find:

(i) f(- 1) 

(ii) f(π/6)

(iii) f(2π/3)

(iv) f(4)

(v) f(6)

Answer 1

(i) f(- 1) = 3^-(-1) - 1

= 3¹ - 1 = 3 - 1 = 2

∴ f(- 1) = 2

(ii) f(π/6) = tan \(\frac{\pi/6}{2}\) = tan \((\frac{\pi}{12})\) [∵ 0 < π / 6/ π]

= tan 15° = 2 - √3

(iii) f \((\frac{2\pi}{3})\) = tan \((\frac{2\pi}{3 \times 2})\) = tan \(\frac{\pi}{3}\) [∵ 0 < 2π/3 < π]

= \(\sqrt3\)

(iv) f(4) = \(\frac{4}{4^2-2} = \frac{4}{16 -2} = \frac{4}{14} = \frac{2}{7}\) = [∵ 0 < (4) < 6]

[∵ x lies in = 4, [π , 6]

(v) f(6) = \(\frac{6}{6^2-2} = \frac{6}{36 -2} = \frac{6}{34} = \frac{3}{17}.\)

[∵ x lies in = 6, [π , 6]