inverse of f(x)=-x^2 +14x+17 x<7

Question

I cant figure out how to get the formula for the inverse function. I also need to include domain i believe x<_

 

Answer 1

y=-x²+14x+17,   x<7. This is a parabola that opens downward with

vertex (7,66). So it will be 1-to-1 you take the branch to the left of x=7.

 

-y=x²-14x-17   completing the square   -y=(x-7)²-49-17=(x-7)²-66  so

(x-7)²=66-y  --> x-7=±√(66-y) -->  x=7 ± √(66-y)   Now if you input

y=66 you get x=7 (the vertex) for  y<66 the  + root gives the right branch

and the  - root gives the left branch.

 

Therefore, since x= f^-1(y)  we have   f^-1(y)=7-√(66-y)   with   y<66.   

Answer 2

Let y = f(x):   y = -x²+14x+17

 

 Complete the square to get vertex form:  

 

     y = -(x²-14x) + 17

 

     y = -(x² - 14x + 49) + 66

 

     y = -(x-7)² + 66

 

       Graph is the left half of a parabola opening downward with vertex (highest point) (7,66).

 

       Range f = (-∞, 66)

 

To find f^-1(x), switch x and y and then solve for y:

 

      x = -(y - 7)² + 66

 

      x - 66 = -(y-7)²

 

     (y - 7)² = 66 - x

 

     y - 7 = ±√(66 - x)

 

          y = 7 ± √(66 - x)

 

Since domain f = range f^-1 (so y < 7) and range f = domain f^-1 (so x < 66) we have:  

 

            f^-1(x) = 7 - √(66 - x), where x < 66 and f^-1(x) < 7

 

Graphically, y = f(x) is the left half of a parabola with vertex (7,66) opening downward and y = f^-1(x) is the bottom half of a parabola with vertex (66,7) opening to the left.