Precal functions

Question

If h(t)=4t²+3t-1 and v(t)=3t+2, calculate h(v^-1(t))

Answer 1

Start by using the fact that v(v^-1(t)) = v^-1(v(t)) = t.

 

3v^-1(t) + 2 = t

 

3v^-1(t) = t - 2

 

v^-1(t) = (t - 2) / 3

 

Now plug this into h(t).

 

h(v^-1(t)) = 4[v^-1(t)]² + 3v^-1(t) - 1

 

= 4[(t - 2) / 3]² + 3[(t - 2) / 3] - 1

 

= 4(t² - 4t + 4)/9 + t - 2 - 1

 

= 4t²/9 - 16t/9 + 16/9 + t - 3

 

= 4t²/9 - 7t/9 - 11/9

Answer 2

First, find v^-1(t):  Let y = v(t) = 3t +2

 

                           Switch t and y to get t = 3y + 2

 

                           Solve for y:  y = (t-2)/3

 

                           So, v^-1(t) = (t-2)/3

 

Thus, h(v^-1(t)) = h[(t-2)/3] = 4[(t-2)/3]² + 3[(t-2)/3] - 1

 

                                             = (4/9)(t² - 4t + 4) + t - 2 - 1

 

                                             = (4/9)t² - (16/9)t + 16/9 + t - 3

 

                                             = (4/9)t² - (7/9)t - 11/9

 

                                             = (1/9)(4t² - 7t - 11)