Question

If $y=f\left(x\right)$ is a 4th degree polynomial function passing through point $(0,1)$ and which increases in the intervals $(1,2),(3,\infty )$ and decreases in the intervals $(-\infty ,1),(2,3)$.
If $f\left(1\right)=-8$, then value of $f\left(2\right)$ is:

• A. $-3$
• B. $-6$
• C. $-20$
• D. $-7$

Answer 1

The correct option is D $-7$

Let the polynomial be $a{x}^4+b{x}^3+c{x}^2+dx+e=f\left(x\right)$

It passes through $(0,1)$

$\therefore e=1$

Given: $f\left(1\right)=-8$

$\Rightarrow a+b+c+d+e=-8$

$\Rightarrow a+b+c+d=-9$

$f^{\prime }\left(x\right)$ is of type $k(x-1)(x-2)(x-3)$

$\therefore 4a{x}^3+3b{x}^2+2cx+d=k(x^3-6x^2+11x-6)$

$\therefore 4a=k\Rightarrow a=\frac{k}{4},3b=-6k\Rightarrow b=-2k$

$\therefore 2c=11k\Rightarrow c=\frac{11k}{2},d=-6k$

$\therefore \frac{k}{4}-2k+\frac{11k}{2}-6k=-9\Rightarrow k=4$

$\therefore a=1,b=-8,c=22,d=-24$

$\therefore f\left(x\right)=x^4-8x^3+22x^2-24x+1$

$\Rightarrow f\left(2\right)=16-8\left(8\right)+22\left(4\right)-24\left(2\right)+1$

$\Rightarrow f\left(2\right)=88+16+1-64-48$

$\Rightarrow f\left(2\right)=-7$