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Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppose that there exists an M>0 s.t. |f'(x)|<=m x,="" ¥="" x€(-1,1)="" a)="" f'="" is="" continuous="" at="" 0="" b)="" f'="" is="" differentiable="" at="" 0="" c)="" ff'="" is="" differentiable="" at="" 0="" d)="" (f')^2="" is="" differentiable="" at="" 0?="" x,="" ¥="" x€(-1,1)="" a)="" f'="" is="" continuous="" at="" 0="" b)="" f'="" is="" differentiable="" at="" 0="" c)="" ff'="" is="" differentiable="" at="" 0="" d)="" (f')^2="" is="" differentiable="" at="">
Most Upvoted Answer
Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppo...
Statement of the problem:
Let f:(-1,1)→R be a differentiable function satisfying f(0)=0. Suppose there exists an M>0 such that |f'(x)|≤M|x| for all x∈(-1,1).
Solution:
Claim 1: f'(x) is continuous at x=0.
Proof:
Let ε>0 be given. We need to find δ>0 such that |f'(x)-f'(0)|<ε for="" all="">
Since f is differentiable, by the mean value theorem, there exists a c∈(0,x) such that f'(c)=f(x)/x. Therefore, |f'(x)-f'(0)|=|f'(c)-f'(0)|=|f(x)/x-f(0)/0|=|f(x)/x|.
Now, for x∈(-1,1)-{0}, we have |f'(x)-f'(0)|=|f(x)/x|. Since |f'(x)|≤M|x|, we get |f(x)/x|≤M for all x∈(-1,1)-{0}.
Therefore, for any ε>0, we can choose δ=1 such that |f'(x)-f'(0)|<ε for="" all="" x∈(-δ,δ)-{0}.="" hence,="" f'(x)="" is="" continuous="" at="" x="">
Claim 2: f'(x) is differentiable at x=0.
Proof:
To show that f'(x) is differentiable at x=0, we need to show that
lim┬(x→0)〖(f'(x)-f'(0))/(x-0) 〗=lim┬(x→0)(f'(x))/x
Since f is differentiable, by the mean value theorem, there exists a c∈(0,x) such that f'(c)=f(x)/x. Therefore, (f'(x)-f'(0))/(x-0)=(f(x)/x-f(0)/0)=f(x)/x.
Now, for x∈(-1,1)-{0}, we have (f'(x)-f'(0))/(x-0)=f(x)/x. Since |f'(x)|≤M|x|, we get |(f'(x)-f'(0))/(x-0)|=|f(x)/x|≤M for all x∈(-1,1)-{0}.
Therefore, lim┬(x→0)〖(f'(x)-f'(0))/(x-0) 〗=lim┬(x→0)(f'(x))/x=0, which implies that f'(x) is differentiable at x=0.
Claim 3: (ff') is differentiable at x=0.
Proof:
Let h(x)=f(x)f'(x). Then,
h'(x)=f'(x)f'(x)+f(x)f''(x)
Since f'(x) and f''(x)
Let f:(-1,1)→R be a differentiable function satisfying f(0)=0. Suppose there exists an M>0 such that |f'(x)|≤M|x| for all x∈(-1,1).
Solution:
Claim 1: f'(x) is continuous at x=0.
Proof:
Let ε>0 be given. We need to find δ>0 such that |f'(x)-f'(0)|<ε for="" all="">
Since f is differentiable, by the mean value theorem, there exists a c∈(0,x) such that f'(c)=f(x)/x. Therefore, |f'(x)-f'(0)|=|f'(c)-f'(0)|=|f(x)/x-f(0)/0|=|f(x)/x|.
Now, for x∈(-1,1)-{0}, we have |f'(x)-f'(0)|=|f(x)/x|. Since |f'(x)|≤M|x|, we get |f(x)/x|≤M for all x∈(-1,1)-{0}.
Therefore, for any ε>0, we can choose δ=1 such that |f'(x)-f'(0)|<ε for="" all="" x∈(-δ,δ)-{0}.="" hence,="" f'(x)="" is="" continuous="" at="" x="">
Claim 2: f'(x) is differentiable at x=0.
Proof:
To show that f'(x) is differentiable at x=0, we need to show that
lim┬(x→0)〖(f'(x)-f'(0))/(x-0) 〗=lim┬(x→0)(f'(x))/x
Since f is differentiable, by the mean value theorem, there exists a c∈(0,x) such that f'(c)=f(x)/x. Therefore, (f'(x)-f'(0))/(x-0)=(f(x)/x-f(0)/0)=f(x)/x.
Now, for x∈(-1,1)-{0}, we have (f'(x)-f'(0))/(x-0)=f(x)/x. Since |f'(x)|≤M|x|, we get |(f'(x)-f'(0))/(x-0)|=|f(x)/x|≤M for all x∈(-1,1)-{0}.
Therefore, lim┬(x→0)〖(f'(x)-f'(0))/(x-0) 〗=lim┬(x→0)(f'(x))/x=0, which implies that f'(x) is differentiable at x=0.
Claim 3: (ff') is differentiable at x=0.
Proof:
Let h(x)=f(x)f'(x). Then,
h'(x)=f'(x)f'(x)+f(x)f''(x)
Since f'(x) and f''(x)
Community Answer
Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppo...
Given:
- Function f: (-1,1) --> R is differentiable.
- f(0) = 0
- There exists M > 0 such that |f'(x)| <= mx="" for="" all="" x="" in="">
To prove:
a) f' is continuous at 0
b) f' is differentiable at 0
c) ff' is differentiable at 0
d) (f')^2 is differentiable at 0
Proof:
Continuity of f' at 0:
- We need to show that the limit of f'(x) as x approaches 0 exists.
- Since f is differentiable, f' exists for all x in (-1,1) and is given that |f'(x)| <=>
- Taking the limit as x approaches 0 on both sides of the inequality, we get:
- lim(x->0) |f'(x)| <= lim(x-="">0) Mx
- |lim(x->0) f'(x)| <=>
- Since M > 0, the only way for the inequality to hold is if lim(x->0) f'(x) = 0.
- Therefore, f' is continuous at 0.
Differentiability of f' at 0:
- We need to show that the limit of [f'(h) - f'(0)] / h as h approaches 0 exists.
- Using the mean value theorem, there exists c in (0,h) such that:
- f'(h) - f'(0) = f'(c) * h
- Taking the absolute value of both sides and dividing by |h|, we get:
- |(f'(h) - f'(0)) / h| = |f'(c)|
- Since |f'(c)| <= mc="" for="" all="" c="" in="" (0,h),="" taking="" the="" limit="" as="" h="" approaches="" 0,="" we="">
- lim(h->0) |(f'(h) - f'(0)) / h| <= lim(h-="">0) Mc = 0
- Therefore, the limit of [f'(h) - f'(0)] / h as h approaches 0 is 0, which implies f' is differentiable at 0.
Differentiability of ff' at 0:
- We need to show that the limit of [f(f'(h)) - f(f'(0))] / h as h approaches 0 exists.
- Using the mean value theorem, there exists c in (0,h) such that:
- f(f'(h)) - f(f'(0)) = f'(f'(c)) * (f'(h) - f'(0))
- Taking the absolute value of both sides and dividing by |h|, we get:
- |(f(f'(h)) - f(f'(0))) / h| = |f'(f'(c)) * (f'(h) - f'(0)) / h|
- Since |f'(f'(c))| <= mf'(c)="" for="" all="" c="" in="" (0,h),="" taking="" the="" limit="" as="" h="" approaches="" 0,="" we="">
- lim(h->0) |(f(f'(h)) - f(f'(0)))
- Function f: (-1,1) --> R is differentiable.
- f(0) = 0
- There exists M > 0 such that |f'(x)| <= mx="" for="" all="" x="" in="">
To prove:
a) f' is continuous at 0
b) f' is differentiable at 0
c) ff' is differentiable at 0
d) (f')^2 is differentiable at 0
Proof:
Continuity of f' at 0:
- We need to show that the limit of f'(x) as x approaches 0 exists.
- Since f is differentiable, f' exists for all x in (-1,1) and is given that |f'(x)| <=>
- Taking the limit as x approaches 0 on both sides of the inequality, we get:
- lim(x->0) |f'(x)| <= lim(x-="">0) Mx
- |lim(x->0) f'(x)| <=>
- Since M > 0, the only way for the inequality to hold is if lim(x->0) f'(x) = 0.
- Therefore, f' is continuous at 0.
Differentiability of f' at 0:
- We need to show that the limit of [f'(h) - f'(0)] / h as h approaches 0 exists.
- Using the mean value theorem, there exists c in (0,h) such that:
- f'(h) - f'(0) = f'(c) * h
- Taking the absolute value of both sides and dividing by |h|, we get:
- |(f'(h) - f'(0)) / h| = |f'(c)|
- Since |f'(c)| <= mc="" for="" all="" c="" in="" (0,h),="" taking="" the="" limit="" as="" h="" approaches="" 0,="" we="">
- lim(h->0) |(f'(h) - f'(0)) / h| <= lim(h-="">0) Mc = 0
- Therefore, the limit of [f'(h) - f'(0)] / h as h approaches 0 is 0, which implies f' is differentiable at 0.
Differentiability of ff' at 0:
- We need to show that the limit of [f(f'(h)) - f(f'(0))] / h as h approaches 0 exists.
- Using the mean value theorem, there exists c in (0,h) such that:
- f(f'(h)) - f(f'(0)) = f'(f'(c)) * (f'(h) - f'(0))
- Taking the absolute value of both sides and dividing by |h|, we get:
- |(f(f'(h)) - f(f'(0))) / h| = |f'(f'(c)) * (f'(h) - f'(0)) / h|
- Since |f'(f'(c))| <= mf'(c)="" for="" all="" c="" in="" (0,h),="" taking="" the="" limit="" as="" h="" approaches="" 0,="" we="">
- lim(h->0) |(f(f'(h)) - f(f'(0)))
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Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppose that there exists an M>0 s.t. |f'(x)|
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Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppose that there exists an M>0 s.t. |f'(x)| for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppose that there exists an M>0 s.t. |f'(x)| covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppose that there exists an M>0 s.t. |f'(x)|.
Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppose that there exists an M>0 s.t. |f'(x)| for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppose that there exists an M>0 s.t. |f'(x)| covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f:(-1,1)-->R be a differentiable function satisfying f(0)=0, suppose that there exists an M>0 s.t. |f'(x)|.
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