Question #338882

For what values of 𝑐 does the curve 𝑓(π‘₯) = 2π‘₯^3 + 𝑐π‘₯^2 + 2π‘₯ have maximum and minimum points?









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.














Suppose 𝑓 is differentiable on ℝ and has two roots. Show that 𝑓′ has at least one root.

1
Expert's answer
2022-05-09T18:20:29-0400

1.


fβ€²(x)=6x2+2cx+2f'(x)=6x^2+2cx+2

Find the critical number(s)


fβ€²(x)=0=>6x2+2cx+2=0f'(x)=0=>6x^2+2cx+2=0

3x2+cx+1=03x^2+cx+1=0x=βˆ’cΒ±c2βˆ’126x=\dfrac{-c\pm\sqrt{c^2-12}}{6}

We consider x∈Rx\in \R

c2βˆ’12β‰₯0=>cβ‰€βˆ’12 or cβ‰₯12c^2-12\ge0=>c\le-\sqrt{12}\ or\ c\ge\sqrt{12}

The curve 𝑓(π‘₯)=2π‘₯3+𝑐π‘₯2+2π‘₯𝑓(π‘₯) = 2π‘₯^3 + 𝑐π‘₯^2 + 2π‘₯ have maximum and minimum for c∈(βˆ’βˆž,βˆ’12)βˆͺ(12,∞).c\in(-\infin, -\sqrt{12})\cup (\sqrt{12}, \infin).


2.



fβ€²(x)=6x2+2cx+2f'(x)=6x^2+2cx+2

Find the critical number(s)


fβ€²(x)=0=>6x2+2cx+2=0f'(x)=0=>6x^2+2cx+2=0

3x2+cx+1=03x^2+cx+1=0x=βˆ’cΒ±c2βˆ’126x=\dfrac{-c\pm\sqrt{c^2-12}}{6}

We consider x∈Rx\in \R

c2βˆ’12β‰₯0=>cβ‰€βˆ’12 or cβ‰₯12c^2-12\ge0=>c\le-\sqrt{12}\ or\ c\ge\sqrt{12}

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=xβˆ’x3.y=x-x^3.

x=ΜΈ0,cβ‰€βˆ’12x\not=0, c\le-\sqrt{12} or cβ‰₯12.c\ge\sqrt{12}.


3.

We say that ff has roots at aa and b,(a<b),b, (a<b), and since ff is continuous and differentiable everywhere as polynomial, it is continuous on [a,b][a, b] and differentiable on (a,b).(a, b). So by the Rolle’s theorem, there is a number kk in (a,b),(a, b), such that fβ€²(k)=0.f'(k)=0. Therefore, kk is a root of fβ€²(x)=0.f'(x)=0.



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