1. x-intercepts

`x+4/x^2=0`

`x^3+4=0`

`x=(-4)^{1//3}~~-1.6`


2. y-intercepts:

We cannot have `x = 0`, so there is NO y-intercept, and there must be an asymptote at `x = 0`.


3. Limits

As x → −∞,

`4/x^2 -> 0` ,

so y → −∞

(In fact, the curve approaches the line y = x on the negative side.)

As x → +∞, y → ∞

(In fact, the curve approaches the line y = x on the positive side.)

 

4. Domain all real x, except `0`

   Range all real y

 

5. Maxima and minima?

`y=x+4/x^2=x+4x^-2`

`(dy)/(dx)=1-8x^-3`

`=1-8/x^3`

Now

`1-8/x^3=0`

when

` x^3-8=0`

That is, when `x=2`

 

So we have a max or min at `(2,3)`.

Now, as x → -∞, `dy/dx → 1`,

and as x → ∞, `dy/dx → 1`.

 

6. Second derivative:

`(d^2y)/(dx^2)=24/x^4>0` for all `x`.

So concave up for all x, except `0`. So `(2,3)` is a MIN.

 

7. Near discontinuity:

As `x → 0^-`, (which means x approaches `0` from the negative side), y → ∞

[Try `x = -1, -0.5, -0.1, -0.01, -0.001` etc to see this].

 

As `x → 0^+`, (`0` from the positive side), y → ∞

[Try `x = 1, 0.5, 0.1, 0.01, 0.001` etc to see this].

So we are ready to sketch the curve:

1234-1-2-3-4246810-2-4xyOpen image in a new page

Graph of `y=x+4/x^2`.

The following features are indicated on the graph:

`x`-intercept `(-1.6,0)` (green dot)

Local minimum `(2,3)` (magenta dot)

Asymptote (magenta dashed line)

Get the Daily Math Tweet!
IntMath on Twitter