+0

# Domain

0
129
2

Let f(x) = x^2 + 2x + 5. If the domain of f is all real numbers, then f does not have an inverse function, but if we restrict the domain of f to an interval [c,\infty), then f may have an inverse function. What is the smallest value of c we can use here, so that f does have an inverse function?

Apr 22, 2021

#1
0

This questions was answered by someone a while back so ill just use there answer "

If we restrict the domain from the vertex to positive infinity then f(x) will be a bijection.

We can read the vertex off as x=-2, so we restict the domain as -2 <= x < +infinity"

Apr 22, 2021
#2
+1

We  can write

y  =  x^2  + 2x  + 5       get x  by itself  by  completing the  square on  x

y - 5  + 1  =    x^2  + 2x + 1

y - 4   =  ( x + 1)^2     take the positive root

sqrt ( y - 4)  =  x + 1

sqrt ( y - 4) - 1  =  x     "swap" x and  y

y =  sqrt ( x - 4)  -  1   =   the "Inverse"

This   function  has  its smallest value  of  -1   when  x  =  4  ...so ( 4 , -1)  is on  the inverse

So  (-1, 4)  is on  f

So  c  =  -1   Apr 22, 2021