Question

21. If f(x)=(x-3)(x+2)(x+5)
(x + 1)(x - 7​

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f(x)=(x-1)(x-3)(x-5)(x-7); f:R->R. Demonstrate that f'(x)=0 has exactly 3 real roots.

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Student Answers

NEELA | STUDENT

f(x) = (x-1)(x-3)(x-5)(x-7)

To show that f'(x) has exactly 3 roots'

We know if f(x) is a continuous function between (a,b) and and f(a) = f(b) Then f'(c) = 0 for c belonging to (a,b) by Roll lls theorem.

So in this case f(x) = (x-1)(x-3)(x-5)(x-7) vanishes for f(1)= f(3)= f(5) = f(7) = 0.

Therfore there exists by Roll's theorem some c1 between (1,3) and some c2 between (3,5) and some c3 between (5,7) for which f'(c1) = f'(c2)= f'(c3) = 0

Or c1 and c2 and c3 are the real 3 didtinct roots.

Step-by-step explanation:

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