Answer:
[tex]C'(t) = \frac{t-3.95}{0.95}[/tex]
Step-by-step explanation:
Given
[tex]C=0.95t+3.95[/tex]
Required
The inverse function
[tex]C=0.95t+3.95[/tex]
Swap C and t
[tex]t=0.95C+3.95[/tex]
Make C the subject
[tex]0.95C = t - 3.95[/tex]
Divide by 0.95
[tex]C = \frac{t}{0.95} - \frac{3.95}{0.95}[/tex]
Hence, the inverse function is:
[tex]C'(t) = \frac{t}{0.95} - \frac{3.95}{0.95}[/tex]
or
[tex]C'(t) = \frac{t-3.95}{0.95}[/tex]
