Question

Please help with calculus

Answer 1

"find the transition points, intervals of increase/decrease, concavity, and asymptotic behavior. Then sketch the graph , with this information indicated.

Answer 2

For the function
\[\large y=x-\frac{ 1 }{ x }\]

Answer 3

Oh and I forgot to include this

Answer 4

Critical points are found where the first derivative is 0, the tangent to the curve at these values is horizontal, try and find the first derivative again

Answer 5

Yes, I believe that is what I did.

Answer 6

or is the dericative 1+x^2?

Answer 7

\[\large y' = 1 + \frac{ 1 }{ x^2 }\]

Answer 8

is the va still 0?

Answer 9

Yes, but you figure that from the original function.

\[y = x-\frac{ 1 }{ x }\]

here x can not be zero, and you can see what happens as you approach zero from both sides.

Answer 10

no, i can not see what happens from my head

Answer 11

It makes no difference to the graph for me..

Answer 12

x can't be zero from the 1/x fraction. No zero in the denominator will be defined.
As you approach zero and get cloaser and closer from the + and - sides...

\[\large \lim_{x \rightarrow 0^+}[x-\frac{ 1 }{ x }]=-\infty \]
\[\large \lim_{x \rightarrow 0^-}[x-\frac{ 1 }{ x }]=+\infty \]

see how the fraction will become larger and larger as x gets smaller and smaller

Answer 13

Aha. What am I doing wrong here?

Answer 14

Although this one is the same answer, the Vertical asmytote is from the original function... y= x-x^-1 , you have it from the first derivative

Answer 15

anyway, you have the derivative now

\[\large y'=1+\frac{ 1 }{ x^2 }\]

The critical points are where that y' is equal to zero. y'=0

Answer 16

\[\large y'=0=1+\frac{ 1 }{ x^2 }\]

You can't solve that for any real value of x, so there are no critical points in this case

Answer 17

Meaning, the function does not change inc/dec anywhere. It is always increasing, or always decreasing.

Answer 18

O right, any value that makes the original function undefined is also a critical point. So x=0 is the only critical point.

Answer 19

The first derivative test for the only critical point x=0, the first derivative is positive on both sides of x=0, so the function is increasing all the time

Answer 20

You good on the crit points and first derivative and how it shows increasing or decreasing?

Answer 21

Yes,

Answer 22

I also have a question. To draw the graph, don't we only need the first OR the second derivative test? Do we have to do both?