Answer:
[tex]a'(d) = \frac{d}{5} + \frac{3}{5}[/tex]
[tex]a(a'(d)) = a'(a(d)) = d[/tex]
Step-by-step explanation:
Given
[tex]a(d) = 5d - 3[/tex]
Solving (a): Write as inverse function
[tex]a(d) = 5d - 3[/tex]
Represent a(d) as y
[tex]y = 5d - 3[/tex]
Swap positions of d and y
[tex]d = 5y - 3[/tex]
Make y the subject
[tex]5y = d + 3[/tex]
[tex]y = \frac{d}{5} + \frac{3}{5}[/tex]
Replace y with a'(d)
[tex]a'(d) = \frac{d}{5} + \frac{3}{5}[/tex]
Prove that a(d) and a'(d) are inverse functions
[tex]a'(d) = \frac{d}{5} + \frac{3}{5}[/tex] and [tex]a(d) = 5d - 3[/tex]
To do this, we prove that:
[tex]a(a'(d)) = a'(a(d)) = d[/tex]
Solving for [tex]a(a'(d))[/tex]
[tex]a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})[/tex]
Substitute [tex]\frac{d}{5} + \frac{3}{5}[/tex] for d in [tex]a(d) = 5d - 3[/tex]
[tex]a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3[/tex]
[tex]a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3[/tex]
[tex]a(a'(d)) = d + 3 - 3[/tex]
[tex]a(a'(d)) = d[/tex]
Solving for: [tex]a'(a(d))[/tex]
[tex]a'(a(d)) = a'(5d - 3)[/tex]
Substitute 5d - 3 for d in [tex]a'(d) = \frac{d}{5} + \frac{3}{5}[/tex]
[tex]a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}[/tex]
Add fractions
[tex]a'(a(d)) = \frac{5d - 3+3}{5}[/tex]
[tex]a'(a(d)) = \frac{5d}{5}[/tex]
[tex]a'(a(d)) = d[/tex]
Hence:
[tex]a(a'(d)) = a'(a(d)) = d[/tex]
