- Thread starter Alyad
- Start date

- Joined
- Nov 24, 2012

- Messages
- 1,534

Hello, and welcome to FMH!I really need help solving this inequality

1/(x-2) + (1/3) * x^3 - 1 > 0

Thanks in advance

We are given to solve:

\(\displaystyle \frac{1}{x-2}+\frac{1}{3}x^3-1>0\)

I would begin by combining all terms on the LHS to get a rational expression from which we may obtain our critical values...can you do this?

- Joined
- Nov 24, 2012

- Messages
- 1,534

What I get is:That is exactly what i have a problem with , the best i got is

(3+(x^3-3)(x-2))(x-2)>0

but im not sure if this is any good. Thanks for replying quickly.

\(\displaystyle \frac{1(3)+x^3(x-2)-1(3)(x-2)}{3(x-2)}>0\)

And this reduces to:

\(\displaystyle \frac{x^4-2x^3-3x+9}{3(x-2)}>0\)

Now, do you find any real roots in the numerator?

- Joined
- Nov 24, 2012

- Messages
- 1,534

Okay, now that I know what level you're at, I would advise you to find the extrema of our numerator. Equate its first derivative to zero, and using Descartes' Rule of Signs, how many real roots are there?I am now, im supposed to find extrema of

f(x)=ln|x-2|+1/12x^4 -x+3

I don't speak english that well, can you just explain how to do this so i can try it on my own?

- Joined
- Nov 24, 2012

- Messages
- 1,534

\(\displaystyle \frac{d}{dx}\left(x^4-2x^3-3x+9\right)=4x^3-6x^2-3\)

There is only 1 sign change, so this means there will be one real root. To determine this one real root, I would use a numeric root finding technique to get an approximation. Are you familiar with the Newton-Raphson method? Depending on your professor, you may simply be allowed to use a CAS (computer algebra system) like WolframAlpha to approximate the root. What do you find?

Normally, one would try to factor or find the 4 roots of the numerator to see if there are additional critical points, in addition to the critical point in the denominator.We are finding the root of the derivative of the quartic, which is a cubic.

Since all 4 roots are imaginary, this leaves only a critical point with the denominator.

Last edited by a moderator:

- Joined
- Nov 24, 2012

- Messages
- 1,534

Using W|A, I find the one real root is approximately:

\(\displaystyle \frac{d}{dx}\left(x^4-2x^3-3x+9\right)=4x^3-6x^2-3\)

There is only 1 sign change, so this means there will be one real root. To determine this one real root, I would use a numeric root finding technique to get an approximation. Are you familiar with the Newton-Raphson method? Depending on your professor, you may simply be allowed to use a CAS (computer algebra system) like WolframAlpha to approximate the root. What do you find?

\(\displaystyle x\approx1.74601665058591\)

Now, if we define:

\(\displaystyle f(x)=x^4-2x^3-3x+9\)

We find:

\(\displaystyle f(1.74601665058591)=2.41003192808928105>0\)

Since \(\displaystyle \lim_{x\to\pm\infty}f(x)=+\infty\)

We may conclude:

\(\displaystyle f_{\min}\approx2.41003192808928105\)

Thus \(f(x)\) has no real roots, and is positive over its entire domain. And so we revisit:

\(\displaystyle \frac{x^4-2x^3-3x+9}{3(x-2)}>0\)

The expression on the LHS will thus have the same sign as:

\(\displaystyle x-2\)

which means the solution for the inequality is:

\(\displaystyle x-2>0\)

or:

\(\displaystyle x>2\)

as you found. But you stated you were trying to find the extrema of:

\(\displaystyle g(x)=\ln|x-2|+\frac{1}{12}x^4-x+3\)

And what we want is critical values, that is values of \(x\) such that \(g'(x)=0\), we as we now know, there are none. This means there are no local extrema, and we must examine the behavior of the function at the boundaries of its domain. Can you give the domain?