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On what interval/s is the following function decreasing? (Use interval notation. Round answers to 2 decimal places.)
m(x) = 4x^3 - 5x^2 - 7x

please explain how you do it, thank you!

Respuesta :

frika

You are given the function [tex]m(x) = 4x^3 - 5x^2 - 7x.[/tex]

1. Find the derivative m'(x):

[tex]m'(x)=4\cdot 3x^2-5\cdot 2x-7=12x^2-10x-7.[/tex]

2. Find stationary points, solving the equation m'(x)=0:

[tex]12x^2-10x-7=0,\\ \\D=(-10)^2-4\cdot 12\cdot (-7)=100+336=436,\\ \\\sqrt{D}=\sqrt{436}=2\sqrt{109},\\ \\x_1=\dfrac{10-2\sqrt{109}}{24}=\dfrac{5-\sqrt{109}}{12},\ x_2=\dfrac{10+2\sqrt{109}}{24}=\dfrac{5+\sqrt{109}}{12}.[/tex]

3. Determine the signs of derivative:

  • when [tex]x<\dfrac{5-\sqrt{109}}{12},[/tex] then [tex]m'(x)>0[/tex] (function is increasing);
  • when [tex]\dfrac{5-\sqrt{109}}{12}\le x\le \dfrac{5+\sqrt{109}}{12},[/tex] then [tex]m'(x)<0[/tex] (function is decreasing);
  • when [tex]x>\dfrac{5+\sqrt{109}}{12},[/tex] then [tex]m'(x)>0[/tex] (function is increasing).

Thus, function is decreasing for  [tex]x\in \left(\dfrac{5-\sqrt{109}}{12}, \dfrac{5+\sqrt{109}}{12}\right)\approx (-0.45,1.29).[/tex]

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