If y= 3x + 6, what is the minimum value of (x^3)(y)?

Answer:Given the statement: if y =3x+6.Find the minimum value of [tex](x^3)(y)[/tex]Let f(x) = [tex](x^3)(y)[/tex]Substitute the value of y ;[tex]f(x)=(x^3)(3x+6)[/tex]Distribute the terms;[tex]f(x)= 3x^4 + 6x^3[/tex]The derivative value of f(x) with respect to x.[tex]\frac{df}{dx} =\frac{d}{dx}(3x^4+6x^3)[/tex]Using [tex]\frac{d}{dx}(x^n) = nx^{n-1}[/tex]we have;[tex]\frac{df}{dx} =(12x^3+18x^2)[/tex]Set [tex]\frac{df}{dx} = 0[/tex]then;[tex](12x^3+18x^2) =0[/tex][tex]6x^2(2x + 3) = 0[/tex]By zero product property;[tex]6x^2=0[/tex]   and 2x + 3 = 0⇒ x=0 and x = [tex]-\frac{3}{2} = -1.5[/tex]then;at x = 0f(0) = 0and  x = -1.5[tex]f(-1.5) = 3(-1.5)^4 + 6(-1.5)^3 = 15.1875-20.25 = -5.0625[/tex]
Hence the minimum value of [tex](x^3)(y)[/tex] is, -5.0625