Question #262420

A polynomials is defined by P(x) = mx³ + mx² + mx + n where m and n are non-zero real constants. Given that x = -n is a root of P(x). Determine the range of the possible values of m.

1
Expert's answer
2021-11-08T16:57:04-0500
P(n)=mn3+mn2mn+n=0P(-n)=-mn^3+mn^2-mn+n=0

n0n\not=0


mn2+mnm+1=0-mn^2+mn-m+1=0

m=1n2n+1m=\dfrac{1}{n^2-n+1}

Let m=f(n)=1n2n+1,mn0m=f(n)=\dfrac{1}{n^2-n+1}, mn\not=0


n2n+1=n2n+14+34n^2-n+1=n^2-n+\dfrac{1}{4}+\dfrac{3}{4}

=(n12)2+34>0,nR=(n-\dfrac{1}{2})^2+\dfrac{3}{4}>0,n\in \R

f(n)=2n1(n2n+1)2f'(n)=-\dfrac{2n-1}{(n^2-n+1)^2}

Find the critical number(s)


f(n)=0=>2n1(n2n+1)2=0=>n=12f'(n)=0=>-\dfrac{2n-1}{(n^2-n+1)^2}=0=>n=\dfrac{1}{2}

If n(,0)(0,12),n\in(-\infin, 0)\cup(0, \dfrac{1}{2}), then f(n)>0,f(n)f'(n)>0, f(n) increases.

If n(12,),n\in(\dfrac{1}{2},\infin), then f(n)<0,f(n)f'(n)<0, f(n) decreases.

The function f(n)f(n) has a local maximum at n=12.n=\dfrac{1}{2}.


f(12)=1(12)212+1=43f(\dfrac{1}{2})=\dfrac{1}{(\dfrac{1}{2})^2-\dfrac{1}{2}+1}=\dfrac{4}{3}

limn0f(n)=limn0+f(n)=1\lim\limits_{n\to0^-}f(n)=\lim\limits_{n\to0^+}f(n)=1

limnf(n)=limnf(n)=0\lim\limits_{n\to-\infin}f(n)=\lim\limits_{n\to\infin}f(n)=0

Then m(0,43]m\in(0, \dfrac{4}{3}]




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