Find all the points of local maxima and minima and the corresponding maximum and minimum values of the function `f(x)=-3/4x^4-8x^3-(45)/2x^2+105` .

Explanation

`f^{prime}(x)=-3 x^{3}-24 x^{2}-45 x`


`=-3 x(x^{2}+8 x+15)=-3 x(x+5)(x+3)`


`f^{prime}(x)=0 Rightarrow x=-5, x=-3, x=0`


`f^{prime prime}(x)=-9 x^{2}-48 x-45`


`=-3(3 x^{2}+16 x+15)`


`f^{prime prime}(0)=-45<0`. Therefore, `x=0` is point of local maxima


`f^{prime prime}(-3)=18>0 .` Therefore, `x=-3` is point of local minima


`f^{prime prime}(-5)=-30<0 .` Therefore `x=-5` is point of local maxima.