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f-x-x-2-2x-6-x-1-x-4-2x-3-2-x-lt-1-show-that-there-is-a-number-c-2-3-such-that-f-c-4-




Question Number 90287 by jagoll last updated on 22/Apr/20
f(x) =  { ((x^2 −2x+6 , x≥1)),((x^4 +2x^3 +2 ,x<1)) :}  show that there is a number   c ∈ (−2,3) such that f(c) = 4
$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\begin{cases}{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{6}\:,\:\mathrm{x}\geqslant\mathrm{1}}\\{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{3}} +\mathrm{2}\:,\mathrm{x}<\mathrm{1}}\end{cases} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{number}\: \\ $$$$\mathrm{c}\:\in\:\left(−\mathrm{2},\mathrm{3}\right)\:\mathrm{such}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{c}\right)\:=\:\mathrm{4} \\ $$
Answered by MJS last updated on 22/Apr/20
x^2 −2x+6=4  x^2 −2x+1=−1  (x−1)^2 =−1  x∉R  x^4 +2x^3 +2=4  x^4 +2x^3 −2=0  g(x)=x^4 +2x^3 −2  g(−2)=−2<0  g(1)=1>0  ⇒ there must be at least one zero in [−2; 1]
$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{6}=\mathrm{4} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}=−\mathrm{1} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} =−\mathrm{1} \\ $$$${x}\notin\mathbb{R} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}=\mathrm{4} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}=\mathrm{0} \\ $$$${g}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2} \\ $$$${g}\left(−\mathrm{2}\right)=−\mathrm{2}<\mathrm{0} \\ $$$${g}\left(\mathrm{1}\right)=\mathrm{1}>\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{there}\:\mathrm{must}\:\mathrm{be}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{zero}\:\mathrm{in}\:\left[−\mathrm{2};\:\mathrm{1}\right] \\ $$

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