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MISCELLANEOUS EXERCISE-8 · Q73

Q.Solve using the intermediate value theorem. Show that x3−5x2+3x+6=0x^3 - 5x^2 + 3x + 6 = 0 has at least two real roots between x=1x=1 and x=5x=5.

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Let f(x)=x3−5x2+3x+6f(x)=x^3-5x^2+3x+6. Being a polynomial, ff is continuous everywhere, in particular on [1,5][1,5].

f(1)=1−5+3+6=5>0f(1)=1-5+3+6=5>0.

f(3)=27−45+9+6=−3<0f(3)=27-45+9+6=-3<0.

f(4)=64−80+12+6=2>0f(4)=64-80+12+6=2>0.

Since ff is continuous on [1,3][1,3] and f(1)>0>f(3)f(1)>0>f(3), the Intermediate Value Theorem gives a root c1∈(1,3)c_1\in(1,3) with f(c1)=0f(c_1)=0. Since ff is also continuous on [3,4][3,4] and f(3)<0<f(4)f(3)<0<f(4), the theorem gives a second root c2∈(3,4)c_2\in(3,4) with f(c2)=0f(c_2)=0. As c1≠c2c_1\ne c_2 (they lie in disjoint intervals), these are two distinct roots, both lying inside [1,5][1,5]. …

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