solvng an equation using a number line: e^x - xe^2 <= 0

[chillintoucan28]

Could someone please help! The "e" is natural log

Factor (e^x)-x(e^x) less than or equal to 0

Then solve using a number line. Thanks!

 

[Deleted member 4993]

Could someone please help! The "e" is natural log

Factor (e^x)-x(e^x) greater than or equal to 0

Can you factor

a - ax ? 0


Then solve using a number line. Thanks!

Please share with us your work/thoughts - so that we know where to begin to help you.

 

[chillintoucan28]

yeah that would be a(1-x), but then what do i do?

 

[stapel]

yeah that would be a(1-x), but then what do i do?

Replace " a " with " e[sup:3li5y019]x[/sup:3li5y019] ". :shock:

Eliz.

 

[Deleted member 4993]

yeah that would be a(1-x), but then what do i do?

Now analyze

If

a(1-x) ? 0

then

either

a ? 0 and (1-x) ? 0

or

a ? 0 and (1-x) ? 0

 

[chillintoucan28]

How do I solve (e^x)-xe^x less than or equal to 0, using a number line? do i factor first?

 

[mmm4444bot]

Hi Chill'n:

... do I factor first?

Yes.

... less than or equal to 0 ...

Since the inequality is stating that the left side is ? zero, it is stating that the sign of the left side is NEGATIVE or ZERO (i.e., never positive).

When you factor, you end up with two factors. These factors each have their own SIGN at any given value of x. You need to analyze ? as Subtohosh states at the other location where you started this discussion. The analysis examines the different posibilities of sign for each factor along the number line. Again, the sign of the left side of the inequality is determined by the product of the factors' signs.

You say that you want to use the number line; that makes me think of building a sign matrix (a chart), to organize the information for analysis. Break the number line into intervals where the factors change sign, then form a column of signs underneath each interval. (I give an example below.)

From the discussion at the other thread you started:

yeah that would be a(1-x), but then what do i do?

Now analyze

If

a(1-x) ? 0

then

either

a ? 0 and (1-x) ? 0

or

a ? 0 and (1-x) ? 0

Subhotosh has represented symbolically the basis for the analysis. (Note that the inequality symbol is switched at the other thread.) Subhotosh is saying that the factors must both be positive OR both be negative in order for their product to be greater than zero. If either one of them is zero, then the product is zero.

I understand that a = e^x in your original posts, but I'll change the definition of this symbol, and give an example sign matrix using the new a(1-x).

Let a = x^2 - 1.

\(\displaystyle \[ \left( \begin{array}{cccc} & x1 & x2 & x3 \\a & + & - & + \\(1-x) & + & + & - \\a(1-x)& + & - & - \end{array} \right)\]\)

The matrix above is my analysis of (x^2 - 1)(1 - x) ? 0.

This matrix shows the number line broken up into three segments. These intervals are labeled x1, x2, and x3.

The rows show the sign for a, for (1-x), and for their product.

The expression a(1-x) is ? 0 in every interval except x1.

The solution set is determined by the boundary between x1 and x2. It turns out that this boundary occurs at x = -1. The solution set is x ? -1.

Here's a sloppy picture that corresponds to the sign matrix.

The row at the bottom shows the sign of a(1-x) in each interval, based on the product of the two signs above it.

Can you relate this method to your problem? Pay attention to the endpoints of each interval (eg: half-open intervals, closed intervals, etc.).

If you need more help after factoring,then please post what you've been able to do so far.

~ Mark

 

[chillintoucan28]

can you actually solve the problem for me, this isnt any type of homework problem, its summer time school doesnt start until sept 2nd....im just cuirous to know how to do it and what the actual answer is

 

[Mrspi]

can you actually solve the problem for me, this isnt any type of homework problem, its summer time school doesnt start until sept 2nd....im just cuirous to know how to do it and what the actual answer is

Um, no. You've been given some very good advice about how to solve this problem.

YOU SHOW US what you've done to try to follow that advice, and we'll comment on your work and correct it if necessary.

 

[chillintoucan28]

ok first let me restate the problem

Factor. Then solve using a number line

this is the equation e^x-xe^x less than or equal to 0

this is what i have done so far:

e^x) - xe^x ? 0

Take the ln of every term...
ln (e^x) - ln( x * e^x) ? 0

The second term can be decomposed based on the product rule...

ln (e^x) - ( ln(x) + ln(e^x) ) ? 0

Now distribute the negative
ln (e^x) - ln(x) - ln(e^x) ? 0

If we combine like terms then the ln(e^x) should cancel...
- ln(x) ? 0

Multiply the negative out to change the direction of the inequality
ln(x) ? 0

Raise both sides by using e as the base
e^ln x ? e^0

x ? e^0
x ? 1

so the number line answer would be
[1, ? )

Is this answer correct, in not what have I done wrong

 

[chillintoucan28]

Since this forum wants me to put an effort out to the problem, here it is...

ok first let me restate the problem

Factor= the equation. Then solve using a number line

this is the equation e^x-xe^x less than or equal to 0

this is what i have done so far:

e^x) - xe^x ? 0

Take the ln of every term...
ln (e^x) - ln( x * e^x) ? 0

The second term can be decomposed based on the product rule...

ln (e^x) - ( ln(x) + ln(e^x) ) ? 0

Now distribute the negative
ln (e^x) - ln(x) - ln(e^x) ? 0

If we combine like terms then the ln(e^x) should cancel...
- ln(x) ? 0

Multiply the negative out to change the direction of the inequality
ln(x) ? 0

Raise both sides by using e as the base
e^ln x ? e^0

x ? e^0
x ? 1

so the number line answer would be
[1, ? )

Is this answer correct, in not what have I done wrong

 

[skeeter]

interesting method ... not incorrect, but interesting.

you could have stopped at the line with
\(\displaystyle \ln{x} \geq 0\)
and said
\(\displaystyle x \geq 1\)
based on the graph of the log function.

I think folks expected you to try this ...

\(\displaystyle e^x - xe^x \leq 0\)
factor ...
\(\displaystyle e^x(1 - x) \leq 0\)

now, the first factor ...
\(\displaystyle e^x\)
is positive for all x. that leaves the other factor
\(\displaystyle (1-x)\)
which equals 0 at x = 1 and is less 0 for values of x greater than 1, hence the solution set
\(\displaystyle [1, \infty)\)

good job.

 

[skeeter]

duplicate post

http://freemathhelp.com/forum/viewtopic.php?f=9&t=29989

 

[stapel]

duplicate post

Threads merged.

 

[mmm4444bot]

Hi Chillin:

How does this interesting (but invalid) method use the number line to solve inequalities?

I ask because you wrote:

solvng an equation using a number line ...

... Then solve using a number line. ...

:?:

~ Mark

 

[chillintoucan28]

now what your supposed to do is factor an equation, lets say x^2-x-6 =(x-3) (x+2) and then show the SOLUTION on the number line, at least that's what I think...

 

[stapel]

...at least that's what I think...

If you don't know what the exercise wants, then I'm afraid it is likely going to be difficult for us to explain how to complete it "correctly". Sorry!

Eliz.

 

[chillintoucan28]

ok, they just want me to factor the equation and then show the values for which the equation is less than or equal to zero ON A NUMBER LINE

 

[Deleted member 4993]

ok, they just want me to factor the equation and then show the values for which the equation is less than or equal to zero ON A NUMBER LINE

In other words, you are supposed find the domain of the function when the range is restricted to positive or zero.

 

[mmm4444bot]

Hi Chill'n:

Click here for Google results on search string [graph interval "number line" notation]

There are many examples of how to show an interval ON A NUMBER LINE.