Question

For function f(x)=x3-6x2+ax+b it is given that f(1)=f(3)=0. Find a and b and hence verify Rolle's theorem on [1,3]

Answer 1

I wish ,
It will be helpful

Answer 2

Given :

$f(x) = x^{3} - 6x^{2} + ax + b$

f(1) = 0  and also :

f(3) = 0

To find :

Values of  and b = ?

To verify :

Rolle's theorem on [1,3].

Solution ;

∴f(1) = f(3)  so :

$1-6 +a +b = 27- 54 +3a +b\\-5 + a + b = -27 + 3a + b \\22=2a$

a = 11

Also f(1) = 0 so :

1 - 6 +11 + b = 0

b = -6

So values of a and b are 11 and -6 respectively.

Now according to Rolle's theorem the derivative of a function f(x) is equal to zero at point c∈ (a,b) if f(x ) is defined in the interval [a,b] , also f(b) should be equal to f(a).

Here it is given that f(1) = f(3) , so Rolle's theorem can be applied :

So , f'(c) = 0

i.e.  :

$3c^{2} - 12c +a = 0\\3c^{2} -12c + 11 = 0$

$c=\frac{12+\sqrt{144-132} }{6}$  or  $\frac{12 - \sqrt{144-132} }{6}$

$c = \frac{12 + \sqrt{12} }{6}$     or    $\frac{12-\sqrt{12} }{6}$

$c = 2+\frac{1}{\sqrt{3} }$   or $2-\frac{1}{\sqrt{3} }$

So we get a point c ∈(2,3) for which f'(x) = 0 ,

Hence Rolle's theorem is verified .