i don't get how to do this cause i dunt think you can factor it
how do u solve this inequality
x^3+3x^2-18x < and equal to 22
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Well $\displaystyle x^3+3x^2-18x-22=0$ has three real roots but none of them are "nice".
Depending on the context I would be tempted to sketch the curve $\displaystyle y=x^3+3x^2-18x-22$ to obtain the approximate roots, and identify the relevant regions where $\displaystyle x^3+3x^2-18x\le 22$
RonL
Hello,
you asked this question already. I couldn't get a simple solution and I can only show you how to get some approximations:
1. $\displaystyle x^3+3x^2-18x \leq 22$ you only can factor the LHS of this inequality:
$\displaystyle x(x-3)(x+6) \leq 22$ but I don't know how to use this "result".
2. What I did:
Consider the function:
$\displaystyle f(x) = x^3+3x^2-18x-22$
Calculate the zeros of f by the Newton method. I got:
$\displaystyle x_1 = -5.534193... \vee x_2 = -1.095279... \vee x_3 = 3.629472...$ Using these values to 2 sign. digits you can factor the term f(x) into:
$\displaystyle f(x) \approx (x+5.53)(x+1.10)(x-3.63)$
Now calculate $\displaystyle f(x) \leq 0$
A product of 3 factors is negative (that's the same as lower zero) if 2 factors have the same sign and one factor is negative. Thus you'll get a system of linear inequalities:
$\displaystyle f(x) \leq 0 \Longrightarrow$
$\displaystyle x+5.53\geq 0 \wedge x+1.1\geq 0 \wedge x-3.63 \leq 0 \; \; \; \vee $
$\displaystyle x+5.53\leq 0 \wedge x+1.1 \leq 0 \wedge x-3.63 \leq 0 \; \; \; \vee $
$\displaystyle x+5.53\geq 0 \wedge x+1.1 \leq 0 \wedge x-3.63\geq 0\; \; \; \vee $
$\displaystyle x+5.53 \leq 0 \wedge x+1.1 \geq 0 \wedge x-3.63 \geq 0$
Now solve these inequalities for x. The final result should be:
$\displaystyle x \leq -5.53 \; \vee \; -1.1 \leq x \leq 3.63$
I've attached diagrams of the graph of $\displaystyle g(x)=f(x)+22=x^3+3x^2-18x$ and $\displaystyle y = 22$. Those parts of the graph which are below y = 22 are marked in green or red.