Question

Find a that so value conclusion of theorem holds for mean f(x)=x² + 1 on [-1,1]​

Answer 1

Answer:

frist take x = ‐1 .

and then in second take 1.

then 1st ans will come 0 and second ans will come 2 .

hope its helps you.

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {x}^{2} + 1 \: \: on \: \: [ - 1, \: 1]$

Step :- 1

Since, f(x) is a polynomial function.

$\bf\implies \:f(x) \: is \: continuous \: on \: [ - 1, \: 1]$

Step :- 2

Given that

$\rm :\longmapsto\:f(x) = {x}^{2} + 1 \:$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}( {x}^{2} + 1)$

$\rm :\longmapsto\:f'(x) = 2x + 0$

$\rm :\longmapsto\:f'(x) = 2x$

Since, f'(x) is a polynomial function

$\bf\implies \:f(x) \: is \: differentiable \: on \: ( - 1,1)$

Since, f(x) is continuous as well as differentiable.

So, Lagranges Mean Value Theorem is applicable.

So, there exist atleast one real number c belongs to (a, b) such that

$\rm :\longmapsto\:f'(c) = \dfrac{f(b) - f(a)}{b - a}$

Now,

Consider,

$\rm :\longmapsto\:f( - 1) = {( - 1)}^{2} + 1 = 1 + 1 = 2$

$\rm :\longmapsto\:f( 1) = {(1)}^{2} + 1 = 1 + 1 = 2$

$\rm :\longmapsto\:f'(c) = 2c$

So, on substituting all these values, we get

$\rm :\longmapsto\:2c = \dfrac{1 - 1}{1 - ( - 1)}$

$\rm :\longmapsto\:2c = 0$

$\bf\implies \:c = 0$

Additional Information :-

Rolle's Theorem

Let f(x) be a real valued function defined on [ a, b ] such that

1. f(x) is continuous on [ a, b ]

2. f(x) is differentiable on ( a, b )

3. f( a ) = f( b )

then there exist atleast one real number c belongs to (a, b) such that f'( c ) = 0.