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Question Number 196183 by cortano12 last updated on 19/Aug/23

Answered by dimentri last updated on 19/Aug/23

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Answered by mr W last updated on 19/Aug/23

f(x)=2x^3 +3x^2 −12x−k f′(x)=6x^2 +6x−12=0 ⇒x^2 +x−2=0 ⇒x_(1,2) =−2,1 f(x_1 )=2(−2)^3 +3(−2)^2 −12(−2)−k=20−k f(x_2 )=2(1)^3 +3(1)^2 −12(1)−k=−7−k such that f(x)=0 has 3 distinct real roots, f(x_1 )f(x_2 )<0 (20−k)(−7−k)<0 ⇒−7<k<20 ⇒answer A)

f(x)=2x3+3x2−12x−kf′(x)=6x2+6x−12=0⇒x2+x−2=0⇒x1,2=−2,1f(x1)=2(−2)3+3(−2)2−12(−2)−k=20−kf(x2)=2(1)3+3(1)2−12(1)−k=−7−ksuchthatf(x)=0has3distinctrealroots,f(x1)f(x2)<0(20−k)(−7−k)<0⇒−7<k<20⇒answerA)

Commented by mr W last updated on 19/Aug/23

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