Limits, Rational roots, and inequalities

[funnytim]

Hi everyone,

I have a few questions about math...this is supposed to be Calculus, even though I see somethings that seem familar from a previous course. But anyway...

I have quite a few questions below, and I really appreciate your time to help me out. This assignment is due in two days (no I didn't leave it until the last minute, but was painstakingly working my way through other questions in the same assignment ), so Any amount of help would be great.


1) Find the rational roots of the polynomial P (x) = 3x^4- x^3- 6x^2+ 5x - 1, then usethis information to solve the following equation 3x^4- x^3- 6x^2+ 5x - 1 = 0.
First, I did P(1) = 3(1)^4- (1)^3- 6(1)^2+ 5(1) - 1 = 0
Then I used short division and obtained (x-1)(3x^3-2x^2-4x+1)=0
Then, P(-1) = 3(-1)^3-2(-1)^2-4(-1)+1 = 0
Short division with the above got me to: (x+1) (3x^2 - 5x +1) .
However, I am not sure where to go from here.

(The answer is supposed to be Rational roots = 1 and 1/3 , Solutions are 1, 1/3, (-1 + (square root 5) / 2 ), (-1 - (square root 5) / 2).

2) Find all values of x that satisfy the following inequality: |-(x + 1)|^2 + 1 > 1.
Here are my steps thus far:
-1 > [-(x+1)^2 + 1] > 1
-1-1 > -(x+1)^2 > 1-1
-2 > -(x+1)^2 > 0
Solving for x gives me x > (square root 2) - 1 , and x > 1 . While x > (square root 2) - 1 is one of the answers, apparantely there should be two more answers, x < -(square root 2) - 1 , and x=-1 . What did I do wrong?

3) Solve for x: e^2x- 2e^x- 1 = 0
I solved for X using the quadratic formula, and obtained 1 +/- (square root 2), however the answer key provides x=ln(1 +/- (square root 2)) . Where did the ln come from?

4) Graph the function f (x) = -x^4+ x^2+ 2 without using the calculator. Indicate x- and y-intercepts, and show the behavior of the function as x approaches +infinity and -infinity.
I first found the derivative of f(x), and obtained -4x^3 +2x +1 . Also found x-int to be +/-(square root 2) and +/- (square root 1) , and y-int was 2. Now am I stumped as to how to proceed.

Lastly,
5) Let f (x) =-x^3,if x < -1
(x + 2)^2, if x > -1

Find the following limits, if they exist: lim as x-->-1- f (x) and
lim as x-->-1+ f (x)
I'm not quite sure how to 'read' the two "let' statements above, and which one to use for which.


Again, thank you very very much for your time and effort.

 

[BigGlenntheHeavy]

\(\displaystyle 3) \ e^{2x}-2e^{x}-1=0 \ \implies \ (e^{x})^{2}-2e^{x}-1=0\)

\(\displaystyle Hence, \ let \ u \ = \ e^{x}, \ then \ we \ have \ u^{2}-2u-1 \ = \ 0, \ \implies \ u \ = \ 1\pm\sqrt2 \ (quadratic \ formula).\)

\(\displaystyle e^x \ is \ always \ > \ 0, \ so \ e^{x} \ = \ 1+\sqrt2 \ \implirs \ x \ = \ ln(1+\sqrt2).\)

\(\displaystyle Check: \ e^{ln(1+\sqrt2)^{2}}-2e^{ln(1+\sqrt2)}-1 \ = \ 0 \ \implies \ (1+\sqrt2)^{2}-2(1+\sqrt2)-1 \ = \ 0\)

\(\displaystyle 1+2\sqrt2+2-2-2\sqrt2-1 \ = \ 0, \ 0 \ = \ 0. \ QED\)

 

[fasteddie65]

For 3x^4- x^3- 6x^2+ 5x - 1 = 0, there are 4 zeros: 1, 0.618.3399..., 1/3, and -1.618034... Two are rational and two are irrational.

 

[fasteddie65]

OOPS! The zeros are 1, 0.61803399..., 1/3, and -1.618034...

 

[fasteddie65]

For |-(x + 1)|^2 + 1 > 1:

|-(x + 1)|^2 > 0

This is true for all real numbers, except x = -1.

 

[fasteddie65]

I misread it. |-(x + 1)|^2 + 1 ? 1 is true for all real numbers.

 

[fasteddie65]

f (x) = -x^4 + x^2 + 2

The first derivative is f'(x) = -4x^3 + 2x.

The y-intercept is 2.

f(x) = -x^4 + x^2 + 2 has 1 sign change, so there is exactly 1 positive zero.

f(-x) = -x^4 + x^2 + 2 also has one sign change, so there is exactly 1 negative zero.

The other two zeros must be complex.

f(x) is even, so the graph is symmetric to the y-axis. Graphing it roughly, the zeros are around ±1.4. Let's try ?2. It works, so x = ±?2 are the real zeros.

Using those two zeros and synthetic division, the remaining zeros are solutions of the depressed equation x^2 + 1 = 0, which means x = ±i.

 

[fasteddie65]

"5) Let f (x) =-x^3,if x < -1
(x + 2)^2, if x > -1

Find the following limits, if they exist: lim as x-->-1- f (x) and
lim as x-->-1+ f (x)"

For the limit as x -> -1-, use the first equation: -(-1)^3 = -(-1) = 1.

For the limit as x --> -1+, use the second: (-1 + 2)^2 = 1^2 = 1.

 

[mmm4444bot]

… this is supposed to be Calculus, even though I see [some things] that seem familar from a previous course … :lol:

You're funny_tim.

 

[mmm4444bot]

… The zeros are 1/3, 1, 0.61803399…, -1.61803399…

Ooh, the golden mean is somehow involved somewhere.

 

[funnytim]

For 3x^4- x^3- 6x^2+ 5x - 1 = 0, there are 4 zeros: 1, 0.618.3399..., 1/3, and -1.618034... Two are rational and two are irrational.

OOPS! The zeros are 1, 0.61803399..., 1/3, and -1.618034...

Sorry, not quite sure how you got that (I don't think I'm allowed calculator, as I have to show work). The answers given are 1, 1/3, [ (-1 +/- (square root 5) ) /2 ]

f (x) = -x^4 + x^2 + 2

The first derivative is f'(x) = -4x^3 + 2x.

The y-intercept is 2.

f(x) = -x^4 + x^2 + 2 has 1 sign change, so there is exactly 1 positive zero.

f(-x) = -x^4 + x^2 + 2 also has one sign change, so there is exactly 1 negative zero.

The other two zeros must be complex.

f(x) is even, so the graph is symmetric to the y-axis. Graphing it roughly, the zeros are around ±1.4. Let's try ?2. It works, so x = ±?2 are the real zeros.

Using those two zeros and synthetic division, the remaining zeros are solutions of the depressed equation x^2 + 1 = 0, which means x = ±i.

I'm in the dark about that 'sign change' you mentioned. Pretty confused about the stuff afterward too.

… this is supposed to be Calculus, even though I see [some things] that seem familar from a previous course … :lol:

You're funny_tim.

uh.....thanks? :S

Thank you for your continued help!

 

[mmm4444bot]

I will respond to your latest questions in a few minutes; three animals are dragging me away fro..m.....t...h..........e..................keyb.......

 

[funnytim]

Re:

I will respond to your latest questions in a few minutes; three animals are dragging me away fro..m.....t...h..........e..................keyb.......

haha ok I appreciate it, need to get this done before I go to bed!
Thanks

 

[mmm4444bot]

… solve the following equation 3x^4- x^3- 6x^2+ 5x - 1 = 0

First, I did P(1) = 3(1)^4- (1)^3- 6(1)^2+ 5(1) - 1 = 0

Then I used short division and obtained (x-1)(3x^3 - 2x^2 - 4x + 1) = 0 This minus sign is a sign error.

3x^3 + 2x^2 - 4x + 1 = 0

Are you familiar with the Rational Root Theorem?

It tells us that the only possible Rational roots for the cubic polynomial factor above are -1, -1/3, 1/3, 1.

I'm looking over your other work …

 

[funnytim]

Re:

… solve the following equation 3x^4- x^3- 6x^2+ 5x - 1 = 0

First, I did P(1) = 3(1)^4- (1)^3- 6(1)^2+ 5(1) - 1 = 0

Then I used short division and obtained (x-1)(3x^3 - 2x^2 - 4x + 1) = 0 This minus sign is a sign error.

3x^3 + 2x^2 - 4x + 1 = 0

Are you familiar with the Rational Root Theorem?

It tells us that the only possible Rational roots for the cubic polynomial factor above are -1, -1/3, 1/3, 1.

I'm looking over your other work …

My bad, you're right, it should be 3x^3 + 2x^2 - 4x + 1 = 0. Unfortunately I'm not too sure how to use the 'rational root theorem' to obtain the answers

 

[mmm4444bot]

… My bad, you're right, it should be 3x^3 + 2x^2 - 4x + 1 = 0. Unfortunately I'm not too sure how to use the 'rational root theorem' to obtain the answers

The exercise is asking for the zeros of P(x).

The overall strategy is to completely factor the fourth-degree polynomial.

Setting each factor equal to 0, allows us to solve for the four zeros of P(x).

The Rational Roots Theorem gives us a way to list all of the possible Rational roots. We test these possibilities until we find one that works.

You started out good. You tested the possible root 1 in the fourth-degree polynomial, and you discovered that 1 is a zero of P(x).

So, (x - 1) is a factor of the fourth-degree polynomial. We divide, as you did, to find another factor.

The other factor is 3x^3 + 2x^2 - 4x + 1.

Again, from the Rational Roots Theorem, we test and find that 1/3 is a root of this cubic factor.

We divide the cubic polynomial by (x - 1/3) to find the quadratic factor.

3x^2 + 3x - 3 = 0

The Quadratic Formula gives us the remaining two zeros of P(x).

 

[mmm4444bot]

(4) Graph the function f (x) = -x^4+ x^2+ 2 without using the calculator. Indicate x- and y-intercepts, and show the behavior of the function as x approaches +infinity and -infinity.

I first found the derivative of f(x), and obtained -4x^3 +2x +1 . Also found x-int to be +/-(square root 2) and +/- (square root 1) , and y-int was 2. Now am I stumped as to how to proceed.

We can use the derivative, in a moment, but ±sqrt(1) are neither zeros of f nor f ', so I'm not following your typing.

Also, since you're only asked to graph f(x), there will be no consideration of any possible imaginary roots.

Here's an outline of what I would do, given this graphing exercise.

f(x) = -x^4 + x^2 + 2

I recognize this as quadratic "in form", so I make a subsitution.

Let z = x^2

-z^2 + z + 2 = 0

The Quadratic Formula gives two solutions: z = -1 or z = 2

Since z is the square of x, it cannot be -1, so that solution gets tossed.

z = 2

x^2 = 2

x = ±sqrt(2)

The x-intercepts are roughly (-1.4, 0) and (1.4, 0). Plot those points.

You got the y-intercept. Good. Plot the point (0, 2) also.

Since I'm familiar with the general shape of a fourth-degree polynomial, I know that a "hump" occurs for some value of x between each x-intercept and the origin.

(Most fourth-degree polynomials have a "W" shape.)

These "humps" represent local maxima, so the first derivative of f tells us the locations.

Solve f '(x) = 0, to find them.

x = ±sqrt(2)/2 (i.e., halfway between the intercepts and the origin)

Now, calculate the height of the graph at these two values of x.

f(sqrt[2]/2) = 9/4

Because f is an even function, it has symmetry about the y-axis.

So, both "humps" occur at a level slightly above the y-intercept. (9/4 is 2.25, right?)

This is enough information for me to sketch a rough graph of f.

Of course, knowing that the leading coefficient of f is negative tells me that the value of f(x) goes to negative infinity when x goes to ±infinity.

Questions?

 

[funnytim]

Drat, I didn't see it went to the 2nd page . Let me read over it and get back to you.

 

[mmm4444bot]

… I'm not too sure how to use the 'rational root theorem' …

Not "too" sure? (Heh, heh)

Sounds like maybe you don't know the theorem at all. (I dunno.)

Well, look it up, if you need to, and ask specific questions, if you want to; although, it's pretty self-explanatory.

 

[funnytim]

Is the rational root theorm the 'thing' we did to find the possible roots in the polynomial, to find what is a zero of P(x)?

Eg P (1) = 3(1)^4 - (1)^3 - 6(1)^2 + 5(1) -1 = 0
I understand how to do the above completely, not sure if thats the name of it lol

edit: just re-read your post..and saw my bad engrish, lol. I'm pretty tired right now (its 2am where I'm at) so excuse my bad grammar...