Graphing Function: (x^2 + x -12)/(x-4)

[Government$]

Homework Statement

I should sketch function (x^2 + x -12)/(x-4).

The Attempt at a Solution

I have problem with first derivative i find it to be (x^2 - 8x +8)/(x-4)^2 with roots at 4 - 2√2, 4 + 2√2, and 4(we lose four because f is not defined at 4). Where at 4 - 2√2 f is at max and at 4 + 2√2 f is at min. But when i plug those two roots in f i find that when x 4 - 2√2 f is = 3.32 and when i plug 4 + 2√2 i find f to be 16.47. How come min is greater then max? Is that even possible?

 

[Dick]

Homework Statement

I should sketch function (x^2 + x -12)/(x-4).

The Attempt at a Solution

I have problem with first derivative i find it to be (x^2 - 8x +8)/(x-4)^2 with roots at 4 - 2√2, 4 + 2√2, and 4(we lose four because f is not defined at 4). Where at 4 - 2√2 f is at max and at 4 + 2√2 f is at min. But when i plug those two roots in f i find that when x 4 - 2√2 f is = 3.32 and when i plug 4 + 2√2 i find f to be 16.47. How come min is greater then max? Is that even possible?

Sure it is possible. They are only a local max and a local min. Start filling in your sketch of the rest of the function. You'll see what's happening.

 

[Government$]

I figured it out by myself when i sketched graph. My professor didn't mention local max and min, he only talked about absolute. Thank you anyway.

 

[tiny-tim]

Hi Government$!

… But when i plug those two roots in f i find that when x 4 - 2√2 f is = 3.32 and when i plug 4 + 2√2 i find f to be 16.47. How come min is greater then max? Is that even possible?

At x = 4 it goes off to infinity.

The graph is effectively two unconnected graphs either side of x = 4, one with a maximum and one with a minimum …

since they're unconnected, there's no reason why the maximum should be more than the minimum!