Question

If $f(x)$ is a cubic polynomial which has local maximum at $x=-1$. If $f(2)=18,f(1)=-1$ and $f^{\prime }(x)$ has local minimum at $x=0$, then

• A. the distance between $(-1,2)$ and $[a,f(a)]$ where $x$ $=a$ is the point of local minima is $2\sqrt{5}$.
• B. $f(x)$ is increasing for $x\in [1,2\sqrt{5}]$.
• C. $f(x)$ has local minima at $x=1$
• D. the value of $f(0)=5$.

Answer 1

Answer: (B), (C)

We have the cubic polynomial function
$f(x)=a{x}^3+b{x}^2+cx+d$
$f(1)=-1=a+b+c+d$
$f(2)=18=8a+4b+2c+d$
$19=7a+3b+c(1)$
Therefore,
$f^{\prime }(x)=3ax2+2bx+c$
$f^{\prime }(-1)=0$
$3a-2b+c=0$
$3a+c=0$
$c=-3a(2)$
and $f^{\prime \prime }(x)=6ax+2b$
$f^{\prime \prime }(0)=0$
$b=0(3)$
Substituting Eqs. (2) and (3) in Eq. (1), we get
$19=7a-3a$
$\Rightarrow a=\frac{19}{4}$
Therefore, $c=-\frac{57}{4}$
and $d=-1-a-b-c$
$=-1-\frac{19}{4}-0-\frac{57}{4}$
$=-\frac{(4+19+57)}{4}=-\frac{80}{4}$
Therefore, $f(x)=\frac{19}{4}{x}^3-\frac{57x}{4}-\frac{80}{4}$
$=\frac{1}{4}\left(19x^3-57x-80\right)$
and $f^{\prime }(x)=0\Rightarrow \frac{1}{4}\left(57x^2-57\right)=0$
$\left(x^2-1\right)=0\Rightarrow x=1,-1$
$f^{\prime }(x)=\frac{57}{4}(x-1)(x+1)$
Therefore, $f^{\prime \prime }(x)=\left(\frac{57}{2}\right)2x$
$\Rightarrow f^{\prime \prime }(1)>0$
The point of minima is $x=1$. Therefore,
$f(1)=\frac{1}{4}(19-57-80)=\frac{59}{2}=\frac{-59}{2}$
Point $\left(1,\frac{-59}{2}\right)$ distance from point $(-1,2)$ is
$\sqrt{(1+1{)}^2{\left(\frac{-59}{2}-2\right)}^2}=\sqrt{4+\frac{(63{)}^2}{4}}$
$=\frac{1}{2}\sqrt{63^2+4^2}$
Hence, $f(x)$ is increasing, that is, $x\ge 1$.