Question 3: If f : [– 5, 5] → R is a differentiable function and if f ′(x) does not vanish anywhere, then prove that f (– 5) ≠ f (5).
Class 12 - Math - Continuity and Differentiability
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Here f(x) is given to be differentiable on [-5,5] and hence on(-5,5)
We know that if a function is differentiable on an interval then it is also continuous on that interval
f(x) is continuous on [-5,5]
Thus all the three conditions of Rolle's
theorem are satisfied and there must exist at least one c ¢ (-5,5) such that f'(c) = 0
Therefore f'(x) vanishes at some c € ( -5,5) . which is not possible according to the given problem . Thus our supposition that
f(-5) = f(5) is wrong .
hence f(-5) ≠ f(5)
We know that if a function is differentiable on an interval then it is also continuous on that interval
f(x) is continuous on [-5,5]
Thus all the three conditions of Rolle's
theorem are satisfied and there must exist at least one c ¢ (-5,5) such that f'(c) = 0
Therefore f'(x) vanishes at some c € ( -5,5) . which is not possible according to the given problem . Thus our supposition that
f(-5) = f(5) is wrong .
hence f(-5) ≠ f(5)
Answered by
2
★ CONTINUITY AND DIFFERENTIABILITY ★
Rolle's theorem verification is the approach by which the respective function can be proved according to the given relation f ( 5 ) ≠ f ( - 5 )
f( x ) is continuous , and hence differentiation can be operated consequently , on the range limit defined as [ - 5 , 5 ]
Given function vanishes for the range limit in ( -5 , 5 ) , clearly not possible , Two given functions , from which one is continuous and other isn't can't be evaluated ,it's same like even = odd , which isn't possible , HENCE , f ( -5 ) ≠ f ( x )
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
Rolle's theorem verification is the approach by which the respective function can be proved according to the given relation f ( 5 ) ≠ f ( - 5 )
f( x ) is continuous , and hence differentiation can be operated consequently , on the range limit defined as [ - 5 , 5 ]
Given function vanishes for the range limit in ( -5 , 5 ) , clearly not possible , Two given functions , from which one is continuous and other isn't can't be evaluated ,it's same like even = odd , which isn't possible , HENCE , f ( -5 ) ≠ f ( x )
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
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