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?
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"
+128475 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
