For what values of π does the curve π(π₯) = 2π₯^3 + ππ₯^2 + 2π₯ have maximum and minimum points?
Show that the minimum and maximum points of every curve in the family of polynomials
π(π₯) = 2π₯^3 + ππ₯^2 + 2π₯ lie on the curve π¦ = π₯ β π₯^3.
Suppose π is differentiable on β and has two roots. Show that πβ² has at least one root.
1
Expert's answer
2022-05-09T18:20:29-0400
1.
fβ²(x)=6x2+2cx+2
Find the critical number(s)
fβ²(x)=0=>6x2+2cx+2=0
3x2+cx+1=0x=6βcΒ±c2β12ββ
We consider xβR
c2β12β₯0=>cβ€β12βorcβ₯12β
The curve f(x)=2x3+cx2+2x have maximum and minimum for cβ(ββ,β12β)βͺ(12β,β).
2.
fβ²(x)=6x2+2cx+2
Find the critical number(s)
fβ²(x)=0=>6x2+2cx+2=0
3x2+cx+1=0x=6βcΒ±c2β12ββ
We consider xβR
c2β12β₯0=>cβ€β12βorcβ₯12β
In this case
xξ =0,cx2=β3x3βx
Substitute
y=2x3β3x3βx+2x
y=βx3+x
Then the minimum and maximum points of every curve in the family of polynomials f(x)=2x3+cx2+2x lie on the curve y=xβx3.
xξ =0,cβ€β12β or cβ₯12β.
3.
We say that f has roots at a and b,(a<b), and since f is continuous and differentiable everywhere as polynomial, it is continuous on [a,b] and differentiable on (a,b). So by the Rolleβs theorem, there is a number k in (a,b), such that fβ²(k)=0. Therefore, k is a root of fβ²(x)=0.
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