Curve sketching: show y' = 15x^4 - 30x^2 + 45 is never 0

[nitro_nay]

Ok, I have two questions here. The first is pretty simple, I think. I have an equation ( y' = 15x^4 - 30x^2 + 45). I know that the equation will never equal zero but how do I prove it mathematically? In other words, I wanted to find the roots, there were none, how do I show my work?

Question two, how do I find the the asymtote of (x^3 - 1)/(x + 2)? Im pretty sure it involves synthetic division and Im pretty sure its a slant, but I dont know exactly how to find it. Any help would be appreciated. Thanks

 

[galactus]

Is that a derivative?. y'?. Are you just wanting to find the roots of

\(\displaystyle \L\\15x^{4}-30x^{2}+45=x^{4}-2x^{2}+3\)

If so, they're all complex.

As for the last problem, you have a vertical asymptote. Where is division by 0?.

An oblique asymptote occurs when the degree of the numerator is one more than the degree of the denominator. You have 2 more.
That means when you long divide you get:

\(\displaystyle \L\\\underbrace{\frac{-9}{x+2}}_{\text{this goes to 0\\ as x gets larger}}+x^{2}-2x+4\)

Which means your graph approaches the parabola \(\displaystyle \L\\x^{2}-2x+4\) asymptotically as \(\displaystyle x\to\infty\).

 

[stapel]

how do I find the the asymtote of (x^3 - 1)/(x + 2)? Im pretty sure it involves synthetic division and Im pretty sure its a slant, but I dont know exactly how to find it.

Actually, there's a vertical asymptote, and no horizontal nor slant asymptote. :shock:

To learn about asymptotes and how to find them (which was supposed to have been covered back in algebra), please study at least two lessons from the following:

. . . . .Google results for "asymptotes vertical horizontal slant"

Once you have familiarized yourself with the basic terms and techniques, please attempt the exercise. If you get stuck, or are unsure of your answers, please reply showing all of your work and reasoning. :idea:

Thank you!

Eliz.

 

[nitro_nay]

Is that a derivative?. y'?. Are you just wanting to find the roots of

\(\displaystyle \L\\15x^{4}-30x^{2}+45=x^{4}-2x^{2}+3\)

If so, they're all complex.

That is a derivative. My original formula was y=3x^5 - 10x^3 + 45x
I was supposed to find minimums and maximums, so I took the first derivative ( y'=15x^4 - 30x^2 +45) and set y' equal to 0, so essentially I am trying to find the roots, because that will give me my minimums and maximums. In this equation though I know there are no minimums or maximums, only inflection points. I just want to know if there is a way to show mathematically that there are no roots for the derivative because math teachers can get picky about showing your work on tests

Thanks

 

[galactus]

You can show that the graph doesn't cross the x-axis.

You can show two complex solutions of the equation. Since there are two, there has to be four because their conjugates must be included.

 

[nitro_nay]

Alrighty, my teacher said to use newton's method. When I did it the answers were very inconsistent (ranging between 1.54 and -6.21, but not steadily decreasing), which I am assuming proves that there is no root

 

[Deleted member 4993]

15x^4 - 30x^2 + 45 = 0

x^4 - 2x^2 + 3 = 0

substitute

u = x^2

u^2 - 2u + 3 = 0

for the equation above

b^2 - 4ac = -8 < 0

hence roots of the equation above and the original equations are complex.