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  • Verify Rolle’s theorem for the function f (x)= x ^ 2 + 2x - 8, x [- 4, 2].

Q1  Verify Rolle’s theorem for the functionf (x) = x^2 + 2x - 8, x \epsilon [- 4, 2].

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According to Rolle's theorem function must be
a )  continuous in given closed interval say [x,y] 
b ) differentiable in given open interval say (x,y)
c ) f(x) = f(y)
Then their exist a c \ \epsilon \ (x,y)  such that   f^{'}(c)= 0
If all these conditions are satisfies then we can verify  Rolle's theorem 
Given function is 
 f (x) = x^2 + 2x - 8 
Now, being a polynomial function,  f (x) = x^2 + 2x - 8 is both continuous in [-4,2] and differentiable in (-4,2)
Now,
f (-4) = (-4)^2 + 2(-4) - 8= 16-8-8=16-16=0
Similalrly,
f (2) = (2)^2 + 2(2) - 8= 4+4-8=8-8=0
Therefore, value of f (-4) = f(2)=0 and  value of f(x) at -4 and 2 are equal
Now,
According to roll's theorem their is point c , c \ \epsilon (-4,2) such that  f^{'}(c)=0
Now,
f^{'}(x)=2x+2\\ f^{'}(c)=2c+2\\ f^{'}(c)=0\\ 2c+2=0\\ c = -1
And   c = -1 \ \epsilon \ (-4,2)
Hence, Rolle's theorem is verified for the given function  f (x) = x^2 + 2x - 8

Posted by

Gautam harsolia

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