The function $f(x)=\frac{-3}{4}x^4-8x^3-\frac{45}{2}x^2 + 163$ has a local maxima at

$x=0$ and $x = -5$

The correct answer is Option (3) → $x=0$ and $x = -5$

$f(x)=-\frac{3}{4}x^{4}-8x^{3}-\frac{45}{2}x^{2}+163$

$f'(x)=-3x^{3}-24x^{2}-45x=-3x\bigl(x^{2}+8x+15\bigr)=-3x(x+3)(x+5)$

Critical points: $x=0,\ -3,\ -5$

$f''(x)=-9x^{2}-48x-45$

$f''(0)=-45<0\ \Rightarrow$ local maximum at $x=0$

$f''(-3)= -9(9)-48(-3)-45=-81+144-45=18>0\ \Rightarrow$ local minimum at $x=-3$

$f''(-5)= -9(25)-48(-5)-45=-225+240-45=-30<0\ \Rightarrow$ local maximum at $x=-5$

Local maxima occur at $x=0$ and $x=-5$. (Local minimum at $x=-3$.)