Solve for x, ER
-x^3+25x < 0
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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.
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)
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