Solve this inequality: $x^4 - 3x^3 - 4x^2 \leq 0$
My answer (wrong) :
$x^2 (x-4)(x+1)\leq0$
Case 1
$x^2 \geq 0$ , $x\geq4$ , $x\leq-1$ NO
Case 2
$x^2 \geq 0$, $x\leq4$ , $x \geq -1$ ... so this is $x>0$, $x<4$
Case 3
$x^2\leq 0$ , $x\geq 4$ , $x\geq -1$ NO
Case 4
$x^2\leq 0$ , $x\leq 4$ , $x\leq-1$ so this is $x\leq -1$
So i got $x\leq-1$ and $4\geq x\geq 0$
The real correct answer is $-1\leq x\leq 4 $
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$\begingroup$$$ x^2(x-4)(x+1) \leq 0 $$ this was correct but your analysis was slightly wrong.
for real values $x^2\geq 0$ always so we can ignore that here.
we need $$ (x-4)(x+1) \leq 0 $$ thus if $(x-4)\leq 0\implies x\leq 4$ $$ x+1 \geq 0\implies x\geq -1 $$ thus $$ -1\leq x \leq 4 $$
$\endgroup$ 5 $\begingroup$The method of examining cases was the way I learned how to do these inequalities when I was in high school (in a now-distant century...). That method is perfectly fine, but I've found it sometimes confuses students when there are multiple factors.
For an improper inequality such as this (or even when it is proper), we can look for the zeroes of the function first:
$$ x^2 \ (x + 1) \ (x - 4) \ = \ 0 \ \ \Rightarrow \ \ x \ = \ -1 \ , \ 0 \ , \ 4 \ \ . $$
The zeroes "partition" the real numbers into four intervals, which can be examined individually: $$ \ x \ < \ -1 \ \ , \ \ -1 \ < \ x \ < 0 \ \ , \ \ 0 \ < \ x \ < \ 4 \ \ , \ \ x > 4 \ \ . $$
Some textbooks suggest calculating a "test number" in each interval, but we can really be more efficient than that: we can instead just look at the sign of the product of these factors. As everyone here has noted, the first factor is always positive except at the function's zeroes, so we only need to look at the factors $ \ (x + 1) \ \ \text{and} \ \ (x - 4) \ $ . The first of these is negative for $ \ x \ < \ -1 \ $ and positive otherwise, the second factor is negative for $ \ x \ < \ 4 \ $ and positive otherwise. So the signs of the three factors and of the product in each interval are
x < -1 : $ \ + \ \cdot \ - \ \cdot \ - \ \ = \ \ + \ \ $ ;
-1 < x < 0 : $ \ + \ \cdot \ + \ \cdot \ - \ \ = \ \ - \ \ $ ;
0 < x < 4 : $ \ + \ \cdot \ + \ \cdot \ - \ \ = \ \ - \ \ $ ;
x > 4 : $ \ + \ \cdot \ + \ \cdot \ + \ \ = \ \ + \ \ $ .
(Crossing successive partitions then becomes like "flipping a binary switch" when the linear factors are raised to odd integer powers, or not doing so if the power is an even integer.)
The proper inequality $ \ x^4 \ - \ 3x^3 \ - \ 4x^2 \ < \ 0 \ $ is thus satisfied in the intervals $ \ -1 \ < \ x \ < \ 0 \ \ \text{and} \ 0 \ < \ x \ < \ 4 \ $ . Since we are working with an improper inequality, we must also include the "zeroes" of the function, which "binds" these intervals together and affixes endpoints. The solution interval is thus $ \ -1 \ \le \ x \ \le \ 4 \ $ .
$\endgroup$ 8 $\begingroup$we have $$x^2(x-4)(x+1)\le 0$$ since $$x^2\geq 0$$ for all real $x$ we obtain the two cases $$x\geq 4$$ and $$x\le -1$$ which is senseless thus we get $$-1\le x\le 4$$
$\endgroup$ $\begingroup$$x^4-3x^3-4x^2\le0\iff x= x^2(x^2-3x-4)\le 0\iff x^2-3x-4\le0$.
Now the quadratic polynomial $ x^2-3x-4\le0$ has two (integer) roots $-1,4$. It's a basic result on quadratic polynomials that the have the sign of the dominant coefficient outside the interval of the roots, the opposite sign between the roots, if any: $$x^2-3x-4\le0\iff -1\le x\le 4.$$
$\endgroup$ $\begingroup$I think you've begun well. x^2(x-4)(x+1)<=0 Since x^2 is always greater than 0, Therefore, it reduces to : (x-4)(x+1) <=0, Which clearly has a solution:
-1<= x <= 4
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