Verify Rolle’s theorem for the function f (x) = (x – a)m (x – b)n on the interval [a, b], where m, n are positive integers.

Asked by Topperlearning User | 5th Aug, 2014, 08:38: AM

Expert Answer:

We have, f (x) = (x – a)m (x – b)n where m, n Î N

On expanding (x – a)m and (x – b)n by binomial theorem and then taking the product, we find that f (x) is a polynomial of degree (m + n). Since a polynomial function is everywhere differentiable and so continuous also. Therefore,
(i)                 f (x) is continuous on [a, b]
(ii)               f (x) is derivable on (a, b)
Also, f (a) = f (b) = 0
Thus, all the three conditions of Rolle’s theorem are satisfied.
Now, we have to show that there exists c Î (a,b) such that f ¢ (c) = 0
We have,
f (x) = (x – a)m (x – b)n
Þ f ¢ (x) = m (x – a)m – 1 (x – b)n + (x – a)m n (x – b)n – 1
Þ f ¢ (x) = (x – a)m – 1 (x – b)n – 1 {m (x – b) + n (x – a)}
Þ f ¢ (x) = (x – a)m – 1 (x – b)n – 1 {x (m + n) – (mb + na)}
\ f ¢ (x) = 0
Þ (x – a)m – 1 (x – b)n – 1{x (m + n) – (mb + na)} = 0
Þ (x – a) = 0 or (x – b) = 0 or x (m + n) – (mb + na) = 0
Þ x = a or, x = b or, xfraction numerator mb plus na over denominator straight m plus straight n end fraction
Since x = fraction numerator mb plus na over denominator straight m plus straight n end fraction divides (a, b) into the ratio m : n. Therefore, fraction numerator mb plus na over denominator straight m plus straight n end fraction Î (a, b).
Thus, c = fraction numerator mb plus na over denominator straight m plus straight n end fraction Î (a, b) such that f ¢ (c) = 0
Hence, Rolle’s theorem is verified.

Answered by  | 5th Aug, 2014, 10:38: AM