Question

If `h ( x ) = x + \sqrt { x } ,` what is `h ^ { - 1} ( 6)`?

Answer 1

`h^(-1)(6) = 4`

Explanation:

This can be solved by observation, but here's a full blown solution that calculates `h^(-1)(y)` for any `y`...

First note that the domain of `h(x)` is `[0, oo)`

If `x < 0` then `sqrt(x)` is not Real, so not part of the domain of `h(x)` considered as a Real-valued function.

Note also that `h(x)` is strictly monotonic increasing on `[0, oo)`. We can tell this since both `x` and `sqrt(x)` increase as `x` increases.

So `h^(-1)(y)` is a well defined function with domain equal to the range of `h(x)`.

Let:

`y = h(x) = x + sqrt(x)`

Subtract `x` from both ends to get:

`y-x = sqrt(x)`

Square both sides (noting that this will introduce spurious solutions) to get:

`y^2-2xy+x^2 = x`

[[ Note that any spurious solutions will actually be solutions of: `y-x = -sqrt(x)` ]]

Subtract `x` from both sides and rearrange to find:

`x^2-(2y+1)x+y^2 = 0`

This is a quadratic in `x` of the form:

`ax^2+bx+c = 0`

with `a=1`, `b=-(2y+1)` and `c = y^2`

We can solve this using the quadratic formula to find:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = ((2y+1)+-sqrt((2y+1)^2-4y^2))/2`

`color(white)(x) = ((2y+1)+-sqrt(4y+1))/2`

Note that one of these roots will be the solution of:

`y-x = sqrt(x)`

and the other the solution of:

`y-x = -sqrt(x)`

The latter of these requires a larger value of `x`, so discard the `+` sign in the quadratic formula to get:

`x = ((2y+1)-sqrt(4y+1))/2`

That is:

`h^(-1)(y) = ((2y+1)-sqrt(4y+1))/2`

So:

`h^(-1)(6) = ((2(6)+1)-sqrt(4(6)+1))/2 = (13-sqrt(25))/2 = (13-5)/2 = 4`

Answer 2

x=4.

Explanation:

I use graph of `x+sqrtx` to find x=h(-1)(6). I am thankful to Prof.

George, for his valuable tips that ranges in Socratic graphs can be

changed mechanically.

Here, the question is on finding abscissa x, when the ordinate is 6,

In the graph, h = 6 meets h = x+sqrtx, at x = 4. The point (4, 6) is

marked.

Note solutions like `1/3, sqrt2, e and pi` can be approximated by a

graphical method and this approximation can be used as a starter,

for a befitting numerical iterative method, to improve the accuracy to

any precision.

graph{(x+sqrtx-y)(y-6)((x-4)^2+(y-6)^2-.01)=0 [0,20, 0, 10]}