Question

If $f\left(x\right)=x^2(x+2)+x+3$, then

• A. $f(-3-k)<0$ and $f(-2+k)>0$ for all $k>0$.
• B. $f(-3-k)>0$ and $f(-2+k)<0$ for all $k>0$.
• C. $f\left(x\right)=0$ has a roots $\alpha$ such that $\left[\alpha \right]+3=0$, where $\left[\alpha \right]$ is the greatest integer less than or equal to $\alpha$.
• D. $f\left(x\right)=0$ has exactly one root $\alpha$ such that $\left(\alpha \right)+2=0$, where $\left(\alpha \right)$ is the smallest integer greater than or equal to $\alpha$.

Answer 1

Answer: (A), (C), (D)

$f(-3-k)<0$ and $f(-2+k)>0$ for all $k>0$.
$f\left(x\right)=0$ has a roots $\alpha$ such that $\left[\alpha \right]+3=0$, where $\left[\alpha \right]$ is the greatest integer less than or equal to $\alpha$.
$f\left(x\right)=0$ has exactly one root $\alpha$ such that $\left(\alpha \right)+2=0$, where $\left(\alpha \right)$ is the smallest integer greater than or equal to $\alpha$.

$f\left(x\right)=x^2(x+2)+x+3$
$f\left(x\right)=x^3+2x^2+x+3$
$f^{\prime }\left(x\right)=3x^2+4x+1$
$f^{\prime }\left(x\right)=0$; when $x=-1$ and $x=-\frac{1}{3}$
$f(-3-k)={(-k-3)}^2(-k-3)-k$
If $k>0$, then this term would always be negative.
$f(-2+k)={(k-2)}^2\left(k\right)+k+1$ for $k>0$, this term would always be positive.
$f\left(x\right)=0$, it is clear from $f\left(x\right)$ that roots will be negative
$f\left(0\right)=3;f(-1)=3;f(-2)=1;f(-3)=-9$, so there is at least a root between $(-2)$ and $(-3)$
Suppose it's $\alpha$, then $\left[\alpha \right]=-3$
$\left[\alpha \right]+3=0$, where $[.]$ is the greatest integer function
$\left(\alpha \right)+2=0$, where $(.)$ is the smallest integer function