# Algebra prob

1. Jul 13, 2009

### 3.141592654

1. The problem statement, all variables and given/known data

(A) Graph the functions f(x)=x/2 and g(x)=1+(4/x) together to indentify the values of x for which x/2>1+(4/x).

(B) Confirm your findings in part (A) using algebra.

2. Relevant equations

3. The attempt at a solution

(A) I graphed the equations and found the answer to be (-2,0)U(4,infinity). The answer is given in the book and confirms this.

(B) x/2 > 1+(4/x),

= x > 2+(8/x)

= (x^2) > 2x+8

= (x^2)-2x-8 > 0

= (x-4)(x+2) > 0

So doesn't this imply x/2 > 1+(4/x) on the interval (-infinity, -2)U(4, infinity)? Thanks for clearing this up!

2. Jul 13, 2009

### Coto

You'll have to be very careful here.

When you have something along the lines of:

x > y

What happens if you multiply both sides by a negative number? This is the same problem you have currently, except with your problem, you will have to consider two cases, one case when x is positive, and another when x is negative, (since you're multiplying through by x) and you will have to deal with the inequality for each case.

Last edited: Jul 13, 2009
3. Jul 13, 2009

### Дьявол

Ok, first you need to state here x > 2+(8/x) that x$\neq$0
Then you need to solve (x-4)(x+2) > 0

What should be (x-4) and (x+2) so that the whole equation is > 0 i.e positive?

4. Jul 13, 2009

### Coto

Дьявол, it's true he should also state that $$x \neq 0$$, but like I said above, in order to get the full answer, he will have to consider two separate cases for when both $$x < 0$$ and $$x > 0$$.

5. Jul 13, 2009

### HallsofIvy

Staff Emeritus
The simplest way to solve an inequality like x/2> 1+ 4/x is to look first at the equality x/2= 1+ 4/x. Now, multiplying on both sides by 2x, $x^2= 2x+ 8$ or $x^2- 2x- 8= (x+ 2)(x- 4)= 0$ which has roots -2 and 4.

The point is that the only way "f(x)> g(x)" can change to "f(x)< g(x)" is where "f(x)= g(x)" or at points where one or the other of the functions is not continuous. Here the functions are equal at -2 and 4 and 1+ 4/x is not continuous at x= 0.

If x= -4< -2, x/2= -2 and 1+ 4/x= 1- 1= 0. Yes, -2< 0 so the inequality is satisfied for all x< -2. If x= -1, which is between -2 and 0, x/2= -1/2 and 1+ 4/x= -3. -1/2 is NOT less than -3 so the inequality is NOT satisfied for all x between -2 and 0. You check for x between 0 and 4, and for x> 4.

6. Jul 16, 2009

### Mentallic

Yes, or...

if multiplying by x is a problem because if x is negative, it would switch the inequality sign around. Well, how about multiplying by x2? That value is positive no matter if x is positive or negative so there is no need to switch the sign.

$$\frac{x}{2}>1+\frac{4}{x}$$

$$\frac{x^3}{2}>\frac{x^2(x+4)}{x}$$

$$x^3>2x^2+8x$$

$$x^3-2x^2-8x>0$$

This is a cubic but can be reduced into a quadratic if x is factored out. All you need to do is sketch the cubic and find the intervals that are above the x-axis (greater than 0). You should get the same answer