PRECALCULUS HELP: inverse of a equation

Question

The function
f(x) = −12 x^2 + 6x + 5
is not invertible, however the domain can be restricted so that the function is invertible only on the restricted domain. There are many ways to do this. One way is to only use the set of positive, real numbers of the form [a,∞). Determine a value of a so that it is the smallest possible, positive real number where the function is invertible if a ≤ x < ∞.

Using your value of a determine the domain for which f(x) is invertible. Provide your answer in interval notation

Answer 1

Determine the vertex. The value of a is the value of the x coordinate.

Answer 2

y = -12x² + 6x + 5

 

   = -12(x² - ½x) + 5

 

   = -12(x² - ½x + 1/16 - 1/16) + 5

 

   = -12(x - 1/4)² + 23/4

 

y - 23/4 = -12(x - 1/4)²   The graph is a parabola opening downward                                        with vertex (1/4, 23/4).

 

The graph fails the Horizontal Line Test, so it is not one-to-one.  Therefore, the function is not invertible.  

 

However, if we just use the right half of the parabola (i.e., restrict the domain to the interval [1/4,∞) ), we get a graph that does pass the Horizontal Line Test.  So, the function y = -12x²+6x+5, x≥1/4 is invertible.  

 

Answer:  a = 1/4