(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
[tex] \frac{t + 1}{2} [/tex] = y
Answer for (a): [tex] g^{-1}(t)[/tex] = [tex] \frac{t + 1}{2} [/tex]
(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y: [tex] y =\frac{t - 3}{4} [/tex]
Answer for (b): [tex] h^{-1}(t)[/tex] = [tex] \frac{t - 3}{4} [/tex]
(c)
[tex] g^{-1} ( h^{-1}(t)) = g^{-1} (\frac{t - 3}{4})[/tex]
replace all t's in the [tex] g^{-1}(t)[/tex] equation with [tex] \frac{t - 3}{4} [/tex]
[tex] g^{-1} (\frac{t - 3}{4})[/tex] = [tex] \frac{ \frac{t-3}{4} + 1}{2} [/tex]
= [tex] \frac{ \frac{t-3}{4} + \frac{4}{4}}{2} [/tex] = [tex] \frac{ \frac{t - 3 + 4}{4}}{2} [/tex] = [tex] \frac{ \frac{t + 1}{4}}{2} = \frac{t + 1}{8} [/tex]
Answer for (c): [tex] g^{-1} ( h^{-1}(t))[/tex] = [tex] \frac{t + 1}{8}[/tex]
(d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1
(e)
h(g(t)) = 8t - 1
y = 8 t - 1
t = 8y - 1
t + 1 = 8y
[tex] \frac{t + 1}{8} [/tex] = y
Answer for (e): inverse of h(g(t)) = [tex] \frac{t + 1}{8} [/tex]
