Question

Suppose $f\left(x\right)=x^3+a{x}^2+bx+c$ satisfies $f(–2)=-10$ and takes the extreme value $\frac{50}{27}$ at $x=\frac{2}{3}.$ Then the value of $a+b+c$ is

Answer 1

$f\left(x\right)=x^3+a{x}^2+bx+cf(-2)=-8+4a-2b+c=-10\Rightarrow 4a-2b+c+2=0\cdots \left(1\right)$

$f^{\prime }\left(x\right)=3x^2+2ax+b{f}^{\prime }\left(\frac{2}{3}\right)=0\therefore 3\times \frac{4}{9}+2a\times \frac{2}{3}+b=0\Rightarrow \frac{4}{3}+\frac{4a}{3}+b=0\Rightarrow 4+4a+3b=0\cdots \left(2\right)$

$f\left(\frac{2}{3}\right)=\frac{50}{27}$
This gives $\frac{8}{27}+\frac{4a}{9}+\frac{2b}{3}+c=\frac{50}{27}$
$\Rightarrow 8+12a+18b+27c=50\Rightarrow 12a+18b+27c=42\Rightarrow 4a+6b+9c=14\cdots \left(3\right)$

Solving $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get
$a=-1,b=0$ and $c=2$