Question

Question #686b0

Answer 1

No. The y axis can't be the slope. This would mean that the tangent of `90^o=0`, but tangent `90^o` is undefined.

Explanation:

First we need to differentiate `y=sqrt(x)` to get a gradient function.

`dy/dx x^(1/2)= 1/2x^(-1/2)= 1/(2sqrt((x))`

So our gradient function is:

`y= 1/(2sqrt(x))`

Plugging in `0`:

`y= 1/(2sqrt(0))=` undefined, division by zero.

The slope dosen't exist at `x=0`

Answer 2

Yes; although technically the slope is "undefined"

Explanation:

If `f(x)=sqrt(x)`
then the slope is given by the first derivative:
`color(white)("XXX")f(sqrt(x))/dx=f(x^(1/2))/dx=1/2x^(-1/2)=1/(2sqrt(x))`

When `x=0` this value is undefined (as `xrarr0^+`, the slope `rarr oo`)

This can be seen in a graph of `sqrt(x)`:
graph{sqrt(x) [-0.54, 9.323, -1.845, 3.086]}

as `xrarr0^+` the slope becomes closer and closer to a vertical line.

Answer 3

The slope at `x=0` is not defined.

Explanation:

For the graph of `y=sqrtx`, the line tangent to the graph at the point where `x=0` is the `y`-axis. The `y`-axis is a vertical line.

The definition of slope does not apply to vertical lines.

Some people say that a vertical line has slope equal to infinity. If the person grading your paper is not one of those people, do not say it on anything to be graded. Just say that the slope is not defined or the slope does not exist.
(I don't know why they do not say it is negative infinity. You'll have to ask those who claim the slope is infinity.)

If `f` is defined at `a` and `lim_(xrarra)abs(f'(x)) = oo`, then the tangent line at `(a,f(a))` is the vertical line `x=a`

The graph of `y=sqrtx` is a similar situation.
The graph only exists for `x >= 0` and `lim_(xrarr0^+)abs(f'(x)) = oo`. So the `y`-axis is tangent to the curve.