Find the inverses of each of the functions below algebraically.
a.3y+5x=18
b.h(t)=−4.9(t+3)^2+45.8
To find the inverse of a function algebraically, you need to swap the variables and solve for the new variable.
a. To find the inverse of the function 3y + 5x = 18:
Step 1: Swap the variables:
3x + 5y = 18
Step 2: Solve for y:
3x + 5y = 18
5y = -3x + 18
y = (-3/5)x + 18/5
So, the inverse of the function 3y + 5x = 18 is given by:
y = (-3/5)x + 18/5
b. To find the inverse of the function h(t) = -4.9(t + 3)^2 + 45.8:
Step 1: Swap the variables:
t = -4.9(h + 3)^2 + 45.8
Step 2: Solve for h:
t = -4.9(h + 3)^2 + 45.8
-4.9(h + 3)^2 = t - 45.8
(h + 3)^2 = (t - 45.8)/(-4.9)
h + 3 = ±sqrt((t - 45.8)/(-4.9))
h = ±sqrt((t - 45.8)/(-4.9)) - 3
Therefore, the inverse of the function h(t) = -4.9(t + 3)^2 + 45.8 is given by:
h = ±sqrt((t - 45.8)/(-4.9)) - 3
3y+5x=18
two step process to find the inverse of a linear function
1. interchange the x and y variables
---> 3x + 5y = 18
2. solve the new equation for y
5y = -3x + 18
y = (-3/5)x + 18/5
since the original was in standard form, we could have left the inverse in standard form as 3x + 5y = 18
b)
let s = -4.9t^2 + 45.8
interchange the variables ...
t = -4.9s^2 + 45.8
4.9t^2 = 45.8 - t
t^2 = (45.8 - t)
t = ±√( (45.8-t)/4.9 ) --> notice that this is not a function