Question

Is there any point `(x, y)` on the curve `y=x^(x(1+1/y)), x > 0,` at which the tangent is parallel to the x-axis?

Answer 1

There is no such point, as far as my math goes.

Explanation:

First, let's consider the conditions of the tangent if it is parallel to the `x`-axis. Since the `x`-axis is horizontal, any line parallel to it must also be horizontal; so it follows that the tangent line is horizontal. And, of course, horizontal tangents occur when the derivative equals `0`.

Therefore, we must first start by finding the derivative of this monstrous equation, which can be accomplished through implicit differentiation:
`y=x^(x+x/y)`
`->lny=(x+x/y)lnx`

Using the sum rule, chain rule, product rule, quotient rule, and algebra, we have:
`d/dx(lny)=d/dx((x+x/y)lnx)`
`->dy/dx*1/y=(x+x/y)'(lnx)+(x+x/y)(lnx)'`
`->dy/dx*1/y=(x+x/y)'(lnx)+(x+x/y)(lnx)'`
`->dy/dx*1/y=(1+(x'y-xdy/dx)/y^2)(lnx)+(x+x/y)(1/x)`
`->dy/dx*1/y=lnx+lnx((y-xdy/dx)/y^2)+1+1/y`
`->dy/dx*1/y=lnx+lnx(1/y-(xdy/dx)/y^2)+1+1/y`
`->dy/dx*1/y=lnx+(lnx)/y-(xlnxdy/dx)/y^2+1+1/y`
`->dy/dx*1/y+(xlnxdy/dx)/y^2=lnx+(lnx)/y+1+1/y`
`->dy/dx(1/y+(xlnx)/y^2)=lnx+(lnx)/y+1+1/y`
`->dy/dx((y+xlnx)/y^2)=lnx+(lnx)/y+1+1/y`
`->dy/dx((y+xlnx)/y^2)=(ylnx+lnx+1+y)/y`
`->dy/dx=((ylnx+lnx+1+y)/y)/((y+xlnx)/y^2)`
`->dy/dx=(y(ylnx+lnx+1+y))/(y+xlnx)`

Wow...that was intense. Now we set the derivative equal to `0` and see what happens.
`0=(y(ylnx+lnx+1+y))/(y+xlnx)`
`0=ylnx+lnx+1+y`
`-ylnx-y=lnx+1`
`-y(lnx+1)=lnx+1`
`y(lnx+1)=-(lnx+1)`
`y=(-(lnx+1))/(lnx+1)`
`y=-1`

Interesting. Now let's plug in `y=-1` and see what we get for `x`:
`y=x^(x(1+1/y))`
`-1=x^(x(1+1/-1))`
`-1=x^(x(1-1))`
`-1=x^0`
`-1=1`

Since this is a contradiction, we conclude that there are no points meeting this condition.

Answer 2

There not exists such a tangent.

Explanation:

`y =x^(x (1 + 1/y)) equiv y^{y/(y+1)} = x^x`. Now calling `f(x,y) = x^x-y^{y/(y+1)} = u(x) + v(y) = 0` we have

`df = f_x dx + f_y dy = (partial u)/(partial x) dx + (partial v)/(partial y) dy = 0` then

`dy/dx = -( (partial u)/(partial x))/((partial v)/(partial y)) = (x^x (1 + Log_e(x))(1 + y)^2)/(y^(y/(1 + y)) (1 + y + Log_e(y))) = ((1 + Log_e(x))(1 + y)^2)/ (1 + y + Log_e(y))`

We see that `dy/(dx)=0 -> {y_0 = -1, x_0 = e^{-1}}` but those values must verify:

`f(x,y_0) = 0` and
`f(x_0,y) = 0`

In the first case, `y_0 = 1` we have

`x^x = -1` which is not attainable in the real domain.

In the second case, `x_0 = e^{-1}` we have

`y^{y/(y+1)} = e^{-1}` or
`y/(y+1)log_e y = -1`

but

`y/(y+1)log_e y > -1` so no real solution also.

Concluding, there is not such a tangent.

Answer 3

The answer from Dr, Cawa K, x = 1/e, is precise.

Explanation:

I had proposed this question to get this value precisely. Thanks to

Dr, Cawas for a decisive answer that approves the revelation that

the double precision y' remains 0 around this interval. y is

continuous and differentiable at x = 1/e. As both the 17-sd double

precision y and y' are 0, in this interval around x = 1/e, it was a

conjecture that x-axis touches the graph in between. And now, it is

proved. I think that the touch is transcendental. .