Find $delta(varepsilon)$ function
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Find such $delta(varepsilon): mathbb{R_+} rightarrow mathbb{R_+}$ that for $f: [0, 1] rightarrow mathbb{R}$ and $f=begin{cases}xlog xtext{, if x $ne$ 0}\0text{, if x = 0}end{cases}$ $|x-y|ledelta(varepsilon) Rightarrow |f(x) - f(y)|le varepsilon $. I don't know how to bound $|xlog x - ylog y| le varepsilon$. Please, give me a hint
continuity epsilon-delta
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Find such $delta(varepsilon): mathbb{R_+} rightarrow mathbb{R_+}$ that for $f: [0, 1] rightarrow mathbb{R}$ and $f=begin{cases}xlog xtext{, if x $ne$ 0}\0text{, if x = 0}end{cases}$ $|x-y|ledelta(varepsilon) Rightarrow |f(x) - f(y)|le varepsilon $. I don't know how to bound $|xlog x - ylog y| le varepsilon$. Please, give me a hint
continuity epsilon-delta
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find such $delta(varepsilon): mathbb{R_+} rightarrow mathbb{R_+}$ that for $f: [0, 1] rightarrow mathbb{R}$ and $f=begin{cases}xlog xtext{, if x $ne$ 0}\0text{, if x = 0}end{cases}$ $|x-y|ledelta(varepsilon) Rightarrow |f(x) - f(y)|le varepsilon $. I don't know how to bound $|xlog x - ylog y| le varepsilon$. Please, give me a hint
continuity epsilon-delta
Find such $delta(varepsilon): mathbb{R_+} rightarrow mathbb{R_+}$ that for $f: [0, 1] rightarrow mathbb{R}$ and $f=begin{cases}xlog xtext{, if x $ne$ 0}\0text{, if x = 0}end{cases}$ $|x-y|ledelta(varepsilon) Rightarrow |f(x) - f(y)|le varepsilon $. I don't know how to bound $|xlog x - ylog y| le varepsilon$. Please, give me a hint
continuity epsilon-delta
continuity epsilon-delta
asked Nov 21 at 15:46
user596269
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You don't actually have to bound $|xlog x - ylog y|$
You can prove that $f$ is continuous in $[0,1]$ because $lim_{xrightarrow 0, x>0} f(x) = 0$ and $f(0) = 0$
You also know that $f$ is differentiable in $]0,1[$
you can then prove that $f'$ is continuous on $[0,1]$
Using the mean value theorem
$$exists M>0, forall(x,y)in ]0,1[, M|x-y| < |f(x)-f(y)|$$
You can then chose $delta(epsilon) = M = sup_{xin[0,1]}|f'(x)| = constant$
because you proved that $f'$ is continuous
answered Nov 21 at 21:03
TheD0ubleT
37718
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
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up vote
0
down vote
You don't actually have to bound $|xlog x - ylog y|$
You can prove that $f$ is continuous in $[0,1]$ because $lim_{xrightarrow 0, x>0} f(x) = 0$ and $f(0) = 0$
You also know that $f$ is differentiable in $]0,1[$
you can then prove that $f'$ is continuous on $[0,1]$
Using the mean value theorem
$$exists M>0, forall(x,y)in ]0,1[, M|x-y| < |f(x)-f(y)|$$
You can then chose $delta(epsilon) = M = sup_{xin[0,1]}|f'(x)| = constant$
because you proved that $f'$ is continuous
answered Nov 21 at 21:03
TheD0ubleT
37718
add a comment |
up vote
0
down vote
You don't actually have to bound $|xlog x - ylog y|$
You can prove that $f$ is continuous in $[0,1]$ because $lim_{xrightarrow 0, x>0} f(x) = 0$ and $f(0) = 0$
You also know that $f$ is differentiable in $]0,1[$
you can then prove that $f'$ is continuous on $[0,1]$
Using the mean value theorem
$$exists M>0, forall(x,y)in ]0,1[, M|x-y| < |f(x)-f(y)|$$
You can then chose $delta(epsilon) = M = sup_{xin[0,1]}|f'(x)| = constant$
because you proved that $f'$ is continuous
answered Nov 21 at 21:03
TheD0ubleT
37718
add a comment |
up vote
0
down vote
up vote
0
down vote
You don't actually have to bound $|xlog x - ylog y|$
You can prove that $f$ is continuous in $[0,1]$ because $lim_{xrightarrow 0, x>0} f(x) = 0$ and $f(0) = 0$
You also know that $f$ is differentiable in $]0,1[$
you can then prove that $f'$ is continuous on $[0,1]$
Using the mean value theorem
$$exists M>0, forall(x,y)in ]0,1[, M|x-y| < |f(x)-f(y)|$$
You can then chose $delta(epsilon) = M = sup_{xin[0,1]}|f'(x)| = constant$
because you proved that $f'$ is continuous
answered Nov 21 at 21:03
TheD0ubleT
37718
You don't actually have to bound $|xlog x - ylog y|$
You can prove that $f$ is continuous in $[0,1]$ because $lim_{xrightarrow 0, x>0} f(x) = 0$ and $f(0) = 0$
You also know that $f$ is differentiable in $]0,1[$
you can then prove that $f'$ is continuous on $[0,1]$
Using the mean value theorem
$$exists M>0, forall(x,y)in ]0,1[, M|x-y| < |f(x)-f(y)|$$
You can then chose $delta(epsilon) = M = sup_{xin[0,1]}|f'(x)| = constant$
because you proved that $f'$ is continuous
answered Nov 21 at 21:03
TheD0ubleT
37718
answered Nov 21 at 21:03
TheD0ubleT
37718
answered Nov 21 at 21:03
TheD0ubleT
37718
answered Nov 21 at 21:03
TheD0ubleT
37718
37718
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