Question

The sum of all the local minimum values of the twice differentiable function $f:R\to R$ defined by $f(x)=x^3-3x^2-\frac{3f^{\prime \prime }(2)}{2}x+f^{\prime \prime }(1)$ is :

• A. -22
• B. 5
• C. -27
• D. 0

Answer 1

Answer: (C)

$f(x)=x^3-3x^2-\frac{3}{2}{f}^{\prime \prime }(2)x+f^{\prime \prime }(1)$
$f^{\prime }(x)=3x^2-6x-\frac{3}{2}{f}^{\prime \prime }(2)\dots \dots$ (ii)
$f^{\prime \prime }(x)=6x-6\dots \dots$ (iii)
Now is $3^{\text{rd }}$ equation
$f^{\prime \prime }(2)=12-6=6$
$f^{\prime \prime }(11=0)$
Use (ii)
$f^{\prime }(x)=3x^2-6x-\frac{3}{2}{f}^{\prime \prime }(2)$
$f^{\prime }(x)=3x^2-6x-\frac{3}{2}\times 6$
$f^{\prime }(x)=3x^2-6x-9$
$f^{\prime }(x)=0$
$3x^2-6x-9=0$
$\Rightarrow x=-1\&3$
Use (iii)
$f^{\prime \prime }(x)=6x-6$
$f^{\prime \prime }(-1)=-12<0$ maxima
$f^{\prime \prime }(3)=12>0$ minima.
Use (i)
$f(x)=x^3-3x^2-\frac{3}{2}{f}^{\prime \prime }(2)x+f^{\prime \prime }(1)$
$f(x)=x^3-3x^2-\frac{3}{2}\times 68x+0$
$f(x)=x^3-3x^2-9x$
$f(3)=27-27-9\times 3=-27$