Solve for x, ER

-x^3+25x < 0

----

I tried it out and i got:

x = 0, -5, 5

therefore if x is less than 5, x < 0

if x is greater than 5, x <0

not sure if it's right though.

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- Oct 6th 2009, 06:41 PMagent2421Math Question Help
Solve for x, ER

-x^3+25x < 0

----

I tried it out and i got:

x = 0, -5, 5

therefore if x is less than 5, x < 0

if x is greater than 5, x <0

not sure if it's right though. - Oct 6th 2009, 06:51 PMartvandalay11
You are very close, but yes there is something wrong

You succeeded in finding the points at which the expression=0, now what we do is plug values in between those points into the expression to see if the inequality is satisified

So we have -5,0, and 5

If you draw a number line and plot these points on it, you will see the number line gets divided into 4 regions namely:

x<-5

-5<x<0

0<x<5

5<x

So we have to test values in each of those regions

So let's try -10 for x<-5,

If x=-10, $\displaystyle -x^3+25x=-(-10)^3+25(-10)=-(-1000)-250=1000-250=750>0$

So the inequality fails

Now let's try -1 for -5<x<0

If x=-2, $\displaystyle -x^3+25x=-(-1)^3+25(-1)=-(-1)-25=1-25=-24<0$

So the inequality holds

Let's try 1 for 0<x<5

If x=1, $\displaystyle -x^3+25x=-(1)^3+25(1)=-(1)+25=-1+25=24>0$

So the inequality fails

Let's try 10 for 5<x

If x=10, $\displaystyle -x^3+25x=-(10)^3+25(10)=-(1000)+250=-1000+250=-750$

So the inequality holds

So the answer is where the inequality holds, namely:

-5<x<0

x>5

When stating your answer you want to say

-5<x<0 or x>5

(A lot of people would say "And" not "or" but this is wrong since x cannot be both <0 and >5)

Also note that it doesn't matter what numbers to choose in the intervals (so choose numbers that are easy to work with) - Oct 7th 2009, 12:28 AMGrandad
Hello everyone!

Just a slight variation on artvandalay11's method is to express $\displaystyle -x^3+25x$ as factors:

$\displaystyle -x^3+25x = x(-x^2+25)$

$\displaystyle =x(5+x)(5-x)$

before plugging your 'test' values of $\displaystyle x$.

It makes the arithmetic much easier, because you don't need to work out the*actual value*, just its*sign*. For instance:

$\displaystyle x=-10$ gives $\displaystyle (-10)\times(-5)\times(15)$, which is $\displaystyle >0$ because $\displaystyle (-^{ve})\times(-^{ve})\times(+^{ve})=(+^{ve})$.

Grandad