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lianna [129]
3 years ago
10

How to find (c,d,e)Please do a step by step working.Thank you.

Mathematics
1 answer:
Natasha2012 [34]3 years ago
8 0
(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
                   \frac{t + 1}{2} = y
Answer for (a): g^{-1}(t) =  \frac{t + 1}{2}

(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: y =\frac{t - 3}{4}

Answer for (b): h^{-1}(t) = \frac{t - 3}{4}

(c)
g^{-1} ( h^{-1}(t)) =  g^{-1} (\frac{t - 3}{4})
replace all t's in the g^{-1}(t) equation with \frac{t - 3}{4}
 g^{-1} (\frac{t - 3}{4}) = \frac{ \frac{t-3}{4} + 1}{2}
= \frac{ \frac{t-3}{4} +  \frac{4}{4}}{2} = \frac{ \frac{t - 3 + 4}{4}}{2} = \frac{ \frac{t + 1}{4}}{2} =  \frac{t + 1}{8}
Answer for (c): g^{-1} ( h^{-1}(t)) = \frac{t + 1}{8}

 (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
\frac{t + 1}{8} = y
Answer for (e): inverse of h(g(t)) = \frac{t + 1}{8}
 
























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