If the critical points (extreme values) of the function \( f(x) \) given by \( f(x)=-\displaystyle\frac{3}{4}x^4-ax^3-\frac{bx^2}{2}+15 \) are −3 and −5, then the value of \( f''(-1) \) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −6
\( \displaystyle f'(x)=-3x^3-3ax^2-bx \)
\( \displaystyle =-x\left(3x^2+3ax+b\right) \)
Since −3 and −5 are critical points and are not 0, they are the roots of \( 3x^2+3ax+b=0 \). By the sum and product of roots:
\( \displaystyle -3+(-5)=-\frac{3a}{3}=-a\ \Rightarrow\ a=8 \)
\( \displaystyle (-3)(-5)=\frac{b}{3}\ \Rightarrow\ b=45 \)
Now
\( \displaystyle f''(x)=-9x^2-6ax-b \)
\( \displaystyle =-9x^2-48x-45 \)
\( \displaystyle f''(-1)=-9+48-45=-6 \)