Question

Find all critical numbers and use the Second Derivative Test to determine all local extrema. y = x^2 - (16)/(x)

Answer 1

@satellite73 can u help?

Answer 2

You must mean first derivative test, unless you're also checking for intervals of concavity?

Do you know the derivative of the given function?
\[y=x^2-\frac{16}{x}\]

Answer 3

yea i found he derivatuve its \[2x + \frac{ 16 }{ x^2 } \]

Answer 4

Okay, you know how to find the critical points? Setting \(y'=0,\) etc?

Answer 5

yes

Answer 6

i got x = -2

Answer 7

Good. But keep in mind that \(y'\) is undefined when \(x=0\), so that would be another critical point.

Answer 8

ok

Answer 9

So when applying the derivative test, you have the intervals
\[(-\infty,-2), (-2,0),\text{ and }(0,\infty).\]
Find the sign of the derivative on each of these intervals.

Answer 10

okay how do i do that?

Answer 11

Take any value of x that lies on one of the intervals. For example, take \(x = -1\). This lies in the second interval \((-2,0)\). \[y'(1)=2(-1)+\frac{16}{(-1)^2}=-18<0\]
Since the derivative is negative on this interval, you know that \(y\) is decreasing on this interval.

Answer 12

ok

Answer 13

so so -infi, -2 i should take -1 right?

Answer 14

No, \(-1\) isn't in that interval. You have to pick a number less than \(-2\).

Answer 15

ohh ok

Answer 16

so -3?

Answer 17

\[\large y = x^2 -\frac{16}{x}\]
\[\large \frac{dy}{dx}=2x+\frac{16}{x^2}\]
\[\large \frac{d^2y}{dx^2}=2-\frac{16}{x^3}\]
At S.P's dy/dx=0
\[\large \frac{2x^3+16}{x^2}=0\]
\[\large 2x^3=-16\]
\[\large x^3=-8\]
\[\large x=-2\]
Sub the x-value into the original equation to find the y-coordinate.
\[\large y=(-2)^2-\frac{16}{-2}\]
\[\large y=12\]
Sub the x-value into the second derivative to determine whether it is a maximum or minimum turning point at that particular point.
\[\large y">0\]
\[\large =>Minimum T.P (Turning Point) at (-2, 12)\]
At I.P's y"=0
\[\large \frac{2x^3-16}{x^3}=0\]
\[\large x^3=8\]
\[\large x=2\]
Draw a table to see whether x=2 is an inflexion point.