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The value of c for which Lagrange’s theorem f(x) = |x| in the interval [-1, 1] is
- a)1/2
- b)1
- c)-1/2
- d)non-existent in the interval
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The value of c for which Lagrange’s theorem f(x) = |x| in the in...
For LMVT to be valid on a function in an interval, the function should be continuous and differentiable on the interval
Here,
f(x) = |x| , Interval : [-1,1]
For h>0,
f’(0) = 1
For h<0,
f’(0) = -1
So, the LHL and RHL are unequal hence f(x) is not differentiable at x=0.
In [-1,1], there does not exist any value of c for which LMVT is valid.
Here,
f(x) = |x| , Interval : [-1,1]
For h>0,
f’(0) = 1
For h<0,
f’(0) = -1
So, the LHL and RHL are unequal hence f(x) is not differentiable at x=0.
In [-1,1], there does not exist any value of c for which LMVT is valid.
Most Upvoted Answer
The value of c for which Lagrange’s theorem f(x) = |x| in the in...
Since Fx =|x| therefore it's graph will be considerd only in the x axis but not in y axis and also its been in the closed interval [-1, 1] so other options will be neglected
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Community Answer
The value of c for which Lagrange’s theorem f(x) = |x| in the in...
For LMVT to be valid on a function in an interval, the function should be continuous and differentiable on the interval
Here,
f(x) = |x| , Interval : [-1,1]
For h>0,
f’(0) = 1
For h<0,
f’(0) = -1
So, the LHL and RHL are unequal hence f(x) is not differentiable at x=0.
In [-1,1], there does not exist any value of c for which LMVT is valid.
Here,
f(x) = |x| , Interval : [-1,1]
For h>0,
f’(0) = 1
For h<0,
f’(0) = -1
So, the LHL and RHL are unequal hence f(x) is not differentiable at x=0.
In [-1,1], there does not exist any value of c for which LMVT is valid.
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The value of c for which Lagrange’s theorem f(x) = |x| in the interval [-1, 1] isa)1/2b)1c)-1/2d)non-existent in the intervalCorrect answer is option 'D'. Can you explain this answer?
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