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v(t)=100(1-t/40)^2

 Aug 23, 2017
 #1
avatar+558 
0

Well, let's say y is v(t). It'll make it simpler, or at least... it will for me.  :)

Then the equation is now y=100(1-t/40)^2

To find the inverse you need to switch the y and x, or in this case, y and t.

That would make t=100(1-y/40)^2

Solve for y

 

First divide both sides by 100 to make t/100=(1-y/40)^2

Then square root both sides to get √(t/100)=1-y/40, which simplifies to √(t)/10=1-y/40.

Then subtract 1 to both sides to get  -1+√(t)/10=-y/40.

Next, multiply -1 to both sides to get 1-√(t)/10=y/40

Then multiply 40 to both sides to get 40-4√(t)=y

That is y=40-4√(t)

Which is v(t)=40-4√(t)  because I had used y instead of v(t).

And that's the answer

v(t)=40-4√(t)

Correct me if I'm wrong. laugh

 Aug 23, 2017
 #2
avatar+100529 
+2

 

 

Substituting x for t, this function is not one-to-one....thus....it has no inverse unless we restrict its domain

 

y = 100 ( 1 - t/40 )^2   divide both sides by 100

 

y / 100  = ( 1 - t/40)^2       take both roots of  sides

 

 ±√ (y/100)  = 1 - t/40       multiply both sides by -1

 

 ±√ ( y / 100 )  = t / 40 - 1     add 1 to both sides

 

1  ±  √ ( y / 100 )  = t/40        multiply both sides by 40

 

40  ± 40  √ ( y / 100 )  = t       swap t and y

 

40  ± 40  √ ( t / 100 )  = y    for  y, write  f-1(t)

 

40  ± 40 √ ( t / 100 )  =   f-1(t)

 

Here is the graph : https://www.desmos.com/calculator/lf5zoe7lei

 

If we restrict the original domain to ( -infinity, 40),  the inverse is represented by the orange graph

If we restrict the original domain to (40, infinity), the inverse is represented by the red graph

 

 

cool cool cool

 Aug 23, 2017

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