Answer to Question #338885 in Calculus for Junior

Question #338885

Show that the minimum and maximum points of every curve in the family of polynomials


𝑓(π‘₯) = 2π‘₯^3 + 𝑐π‘₯^2 + 2π‘₯ lie on the curve 𝑦 = π‘₯ βˆ’ π‘₯^3

1
Expert's answer
2022-05-10T06:37:17-0400

fβ€²(x)=6x2+2cx+2f'(x)=6x^2+2cx+2

f"(x)=12x+2cf"(x)=12x+2c

fβ€²(x)=0f'(x)=0

6x2+2cx+2=06x^2+2cx+2=0

x1=βˆ’2c+4c2βˆ’4812x_1=\frac{-2c+\sqrt{4c^2-48}}{12}

x2=βˆ’2cβˆ’4c2βˆ’4812x_2=\frac{-2c-\sqrt{4c^2-48}}{12}

12x(βˆ’2c+4c2βˆ’4812)+2c=012x(\frac{-2c+\sqrt{4c^2-48}}{12})+2c=0

βˆ’2c+4c2βˆ’48=βˆ’2c-2c+\sqrt{4c^2-48}=-2c

4c2βˆ’48=14c^2-48=1

c=49/4=Β±3.5c=\sqrt{49/4}=\plusmn 3.5

If c=3.5

fβ€²(x)=6x2+7x+2=0f'(x)=6x^2+7x+2=0

D=49βˆ’48=1D=49-48=1

x=βˆ’7Β±112x=\frac{-7 \plusmn \sqrt1}{12}

x1=βˆ’0.5x_1=-0.5

x2=βˆ’2/3x_2=-2/3

f"(βˆ’0.5)=12(βˆ’0.5)+2(3.5)=1>0f"(-0.5)=12(-0.5)+2(3.5)=1>0

So, it is minimum

f"(βˆ’2/3)=12(βˆ’2/3)+2(3.5)=βˆ’1<0f"(-2/3)=12(-2/3)+2(3.5)=-1<0

So, it is maximum


If c=-3.5


fβ€²(x)=6x2βˆ’7x+2=0f'(x)=6x^2-7x+2=0

D=49-48=1

x=7Β±112x=\frac{7 \plusmn \sqrt1}{12}

x1=0.5

x2=2/3

f"(βˆ’0.5)=12(0.5)+2(3.5)=13>0f"(-0.5)=12(0.5)+2(3.5)=13>0

So it is minimum

f"(2/3)=12(2/3)+2(3.5)=15>0


So it is minimum


In this case

x=ΜΈ0,cx2=βˆ’3x3βˆ’xx\not=0, cx^2=-3x^3-x

Substitute

y=2x3βˆ’3x3βˆ’x+2xy=2x^3-3x^3-x+2x

y=βˆ’x3+xy=-x^3+x

Then the minimum and maximum points of every curve in the family of polynomials f(x)=2x3+cx2+2xf(x)=2x^3+cx^2+2x lie on the curve y=x-x3





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