P(x)= -0.0013x^3+0.3x^2+8x-372. Find the maximum without graphing. Please show step by step.

Answer:The Maximum value is [tex]P(x)=2076.227[/tex]Step-by-step explanation:Given,[tex]P(x)=-0.0013\times x^{3} +0.3\times x^{2} +8x-372[/tex] (equation-1)Differentiate above equation with respect to 'x',[tex]P'(x)=-0.0039\times x^{2} +0.6x+8[/tex] --- (equation 2)Again differentiate above equation with respect to 'x',[tex]P''(x)=-0.0078\times x +0.6[/tex] ------- (equation 3)From equation-2 we see,The value of [tex]a=-0.0039[/tex] , [tex]b=0.6[/tex] , [tex]c=8[/tex].Now, for maximum or minimum, the first derivative must be 0.For maximum, [tex]P''(x)<0[/tex]So, [tex]P'(x)=-0.0039\times x^{2} +0.6x+8 = 0[/tex]Using the quadratic formula, we find the roots of [tex]P'(x)[/tex] [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac } }{2a}[/tex] [tex]x=\frac{-0.6\pm \sqrt{0.6^{2}-4\times -0.0039\times 8 } }{2\times -0.0039}[/tex][tex]x=\frac{-0.6\pm 0.696}{-0.0078}[/tex][tex]x=-12.3[/tex] or [tex]x=166.15[/tex]For [tex]x=-12.3[/tex],[tex]P''(x)=(6\times-0.0013\times -12.3 )+(2\times 0.3)[/tex][tex]P''(x)=0.696>0[/tex]Which is minimum value at [tex]x=-12.3[/tex]And for [tex]x=166.15[/tex],[tex]P''(x)=(6\times-0.0013\times 166.15 )+(2\times 0.3)[/tex][tex]P''(x)=-0.696< 0[/tex]Which is maximum value at  [tex]x=166.15[/tex]Plug [tex]x=166.15[/tex] in equation-1,[tex]P(x)=-0.0013\times 166.15^{3} +0.3\times 166.15^{2} +8\times 166.15-372[/tex][tex]P(x)=2076.227[/tex]So the Maximum value is [tex]P(x)=2076.227[/tex]