Respuesta :
Answer:
[tex](a)\ y =0\ or\ y=4\ or\ y=6[/tex]
[tex](b) \ (-\infty,4)\ u\ (6,\infty)[/tex]
[tex](c)\ (4,6)[/tex]
Step-by-step explanation:
Given
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
Solving (a): The constants
This implies that:
[tex]\frac{dy}{dt} = 0[/tex]
So, we have:
[tex]y^4 - 10y^3 + 24y^2 = 0[/tex]
Factorize:
[tex]y^2(y^2 - 10y + 24) = 0[/tex]
Expand and Factorize the expression in bracket
[tex]y^2(y^2 - 6y - 4y + 24) = 0[/tex]
[tex]y^2(y(y - 6) - 4(y - 6)) = 0[/tex]
[tex]y^2(y - 4)(y - 6) = 0[/tex]
Split
[tex]y^2 =0\ or\ y-4=0\ or\ y-6=0[/tex]
Solve for y
[tex]y =0\ or\ y=4\ or\ y=6[/tex]
The above represents the constants
Solving (b): Increasing values
Here
[tex]\frac{dy}{dx} > 0[/tex]
Follow the same process in (a) but replace all = with >
i.e.
[tex]y^2(y - 4)(y - 6) > 0[/tex]
[tex]y >0\ or\ y>4\ or\ y>6[/tex]
Test the values of each inequality
For [tex]y > 6[/tex]; say [tex]y = 7[/tex]
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
[tex]\frac{dy}{dt} = 7^4 - 10*7^3 + 24*7^2[/tex]
[tex]\frac{dy}{dt} = 147[/tex]
[tex]147 > 0[/tex]
This is true for [tex]\frac{dy}{dx} > 0[/tex]; This implies that: [tex](6, \infty)[/tex]
For [tex]y > 4[/tex] Say [tex]y = 5[/tex]
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
[tex]\frac{dy}{dt} = 5^4 - 10*5^3 + 24*5^2[/tex]
[tex]\frac{dy}{dt} = -25[/tex]
This is false:
Say [tex]y = 3[/tex] i.e. [tex]y < 4[/tex]
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
[tex]\frac{dy}{dt} = 3^4 - 10*3^3 + 24*3^2[/tex]
[tex]\frac{dy}{dt} = 27[/tex]
[tex]27 > 0[/tex]
This is true for [tex]\frac{dy}{dx} > 0[/tex]. This implies that: [tex](0,4)[/tex]
For [tex]y > 0[/tex]
[tex]y > 0[/tex] implies that: [tex]y < 4[/tex]
For [tex]y < 0[/tex] say [tex]y = -1[/tex]
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
[tex]\frac{dy}{dt} = (-1)^4 - 10*(-1)^3 + 24*(-1)^2[/tex]
[tex]\frac{dy}{dt} = 35[/tex]
This is true for [tex]\frac{dy}{dx} > 0[/tex]. This implies that [tex](-\infty, 0)[/tex]
We have: [tex](-\infty,0), (0,4)\ and\ (6,\infty)[/tex]
[tex](-\infty,0)\ and\ (0,4)[/tex] can be combined to give: [tex](-\infty,4)[/tex]
So, the interval is: [tex](-\infty,4)\ u\ (6,\infty)[/tex]
Solving (c): Decreasing values
Here
[tex]\frac{dy}{dx} < 0[/tex]
i.e.
[tex]y^2(y - 4)(y - 6) < 0[/tex]
[tex]y <0\ or\ y<4\ or\ y<6[/tex]
Test the values of each inequality
For [tex]y < 6[/tex]; say [tex]y = 5[/tex]
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
[tex]\frac{dy}{dt} = 5^4 - 10*5^3 + 24*5^2[/tex]
[tex]\frac{dy}{dt} = -25[/tex]
This is true for [tex]\frac{dy}{dx} < 0[/tex]; This implies that: [tex](\infty,6)[/tex]
For [tex]y < 4[/tex]; say [tex]y = 3[/tex]
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
[tex]\frac{dy}{dt} = 3^4 - 10*3^3 + 24*3^2[/tex]
[tex]\frac{dy}{dt} = 27[/tex]
This is false
Say [tex]y = 5[/tex] i.e. [tex]y > 4[/tex]
This is the same as: [tex]y < 6[/tex]
So, we have the interval to be: [tex](4,\infty)[/tex]
For [tex]y < 0[/tex] say [tex]y = -1[/tex]
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
[tex]\frac{dy}{dt} = (-1)^4 - 10*(-1)^3 + 24*(-1)^2[/tex]
[tex]\frac{dy}{dt} = 35[/tex]
This is false
Say [tex]y = 2[/tex] i.e [tex]y > 0[/tex]
[tex]\frac{dy}{dt} = y^4 - 10y^3 + 24y^2[/tex]
[tex]\frac{dy}{dt} = 2^4 - 10*2^3 + 24*2^2[/tex]
[tex]\frac{dy}{dt} = 32[/tex]
This is also false
So, the intervals are: [tex](4,\infty)[/tex] and [tex](\infty,6)[/tex].
This can be merged as: [tex](4,6)[/tex]
Hence, the interval for [tex]\frac{dy}{dt} < 0[/tex] is [tex](4,6)[/tex]
