For each of the following function, evaluate this expression

$\frac{f(x+h)-f(x)}{h}$

1. $f(x)={x}^{2}-4x+5$

2. $g(x)={x}^{3}-2x+3$

3. $h(x)=\frac{1}{2x}$

$\frac{f(x+h)-f(x)}{h}$

1. $f(x)={x}^{2}-4x+5$

2. $g(x)={x}^{3}-2x+3$

3. $h(x)=\frac{1}{2x}$

Baladdaa9
2022-07-31
Answered

For each of the following function, evaluate this expression

$\frac{f(x+h)-f(x)}{h}$

1. $f(x)={x}^{2}-4x+5$

2. $g(x)={x}^{3}-2x+3$

3. $h(x)=\frac{1}{2x}$

$\frac{f(x+h)-f(x)}{h}$

1. $f(x)={x}^{2}-4x+5$

2. $g(x)={x}^{3}-2x+3$

3. $h(x)=\frac{1}{2x}$

You can still ask an expert for help

akademiks1989rz

Answered 2022-08-01
Author has **16** answers

Step 1

1) $\frac{f(x+h)-f(x)}{h}=\frac{(x+h{)}^{2}-4(x+h)+5-({x}^{2}-4x+5)}{h}$

$=\frac{\text{\u29f8}{x}^{2}+2xh+{h}^{2}-\text{\u29f8}4x-4h+\text{\u29f8}5-\text{\u29f8}{x}^{2}+\text{\u29f8}4x-\text{\u29f8}5}{h}$

$=\frac{h(2x+h-4)}{h}=2x+h-4$

Step 2

2) $\frac{g(x+h)-g(x)}{h}=\frac{(x+h{)}^{3}-2(x+h)+3-({x}^{3}-2x+3)}{h}$

$=\frac{{x}^{3}+3{x}^{2}h+3x{h}^{2}+{h}^{3}-2x-2h+3-{x}^{3}+2x-3}{h}$

$=\frac{h(3{x}^{2}+3xh+{h}^{2}-2)}{h}$

$=3{x}^{2}+3xh+{h}^{2}-2$

Step 3

3) $\frac{h(x+h)-h(x)}{h}=\frac{\frac{1}{2(x+h)}-\frac{1}{2x}}{h}$

$=\frac{2x-2(x+h)}{4xh(x+h)}$

$=\frac{-2h}{4xh(x+h)}=\frac{-1}{2x(x+h)}$

1) $\frac{f(x+h)-f(x)}{h}=\frac{(x+h{)}^{2}-4(x+h)+5-({x}^{2}-4x+5)}{h}$

$=\frac{\text{\u29f8}{x}^{2}+2xh+{h}^{2}-\text{\u29f8}4x-4h+\text{\u29f8}5-\text{\u29f8}{x}^{2}+\text{\u29f8}4x-\text{\u29f8}5}{h}$

$=\frac{h(2x+h-4)}{h}=2x+h-4$

Step 2

2) $\frac{g(x+h)-g(x)}{h}=\frac{(x+h{)}^{3}-2(x+h)+3-({x}^{3}-2x+3)}{h}$

$=\frac{{x}^{3}+3{x}^{2}h+3x{h}^{2}+{h}^{3}-2x-2h+3-{x}^{3}+2x-3}{h}$

$=\frac{h(3{x}^{2}+3xh+{h}^{2}-2)}{h}$

$=3{x}^{2}+3xh+{h}^{2}-2$

Step 3

3) $\frac{h(x+h)-h(x)}{h}=\frac{\frac{1}{2(x+h)}-\frac{1}{2x}}{h}$

$=\frac{2x-2(x+h)}{4xh(x+h)}$

$=\frac{-2h}{4xh(x+h)}=\frac{-1}{2x(x+h)}$

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How to re-write one fraction as two others.

I have the two following fractions.

$\frac{A}{B{x}^{\alpha +1}}$

and

$\frac{C}{D{x}^{\alpha +\beta}}$

The form i want

$\frac{E}{F{x}^{\alpha +\beta +1}}$

I was thinking to do partial fractions or potentially doing a Laplace or Fourier transform, then rearranging and transforming back. How could one possibly get the two fractions into one fraction of this form.\ A,B,C,D,E and F can be number or possible a factorial if we go down the laplace transform route, as can α,β

I have the two following fractions.

$\frac{A}{B{x}^{\alpha +1}}$

and

$\frac{C}{D{x}^{\alpha +\beta}}$

The form i want

$\frac{E}{F{x}^{\alpha +\beta +1}}$

I was thinking to do partial fractions or potentially doing a Laplace or Fourier transform, then rearranging and transforming back. How could one possibly get the two fractions into one fraction of this form.\ A,B,C,D,E and F can be number or possible a factorial if we go down the laplace transform route, as can α,β

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To find:

The ratio of whole numbers using fractional notation in simplest form.

The given ratio is 7.7 to 10

The ratio of whole numbers using fractional notation in simplest form.

The given ratio is 7.7 to 10

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I was reading about limits of data compression and I had this question that kept me thinking.

Alphabet set: Set of all the unique symbols in the input data

Q. If the distribution over the alphabet set is uniform then is it possible to compress data theoretically ? Standard compression methods (like Huffman codes) exploit the bias in the frequency distribution to obtain codes that have a lower average length of the input text. Is it possible to work without frequency?

Idea 1: One idea is that since the distribution is uniform over the symbols, then any assignment of codes would result in the same number of symbols in the range of the mapping, thus we would require the same number of bits to store that information. Hence the average length of compressed data should be greater than or equal to the original average length.

Idea 2: Another idea that pops up in my head is that if the distribution of symbols is uniform, why not create a new symbol set S' that is essentially a set of all the bi-grams of symbols. But, this distribution also needs to be uniform as the probability of a pair of symbols is same for all the possible cases.

These ideas force me to conclude that it is not possible to compress data if the distribution is uniform.

Please correct me if I am wrong at any point, also I would like to know your views on the problem.

Edit: If you can also provide a formal proof or a book for reference then it would be great.

Alphabet set: Set of all the unique symbols in the input data

Q. If the distribution over the alphabet set is uniform then is it possible to compress data theoretically ? Standard compression methods (like Huffman codes) exploit the bias in the frequency distribution to obtain codes that have a lower average length of the input text. Is it possible to work without frequency?

Idea 1: One idea is that since the distribution is uniform over the symbols, then any assignment of codes would result in the same number of symbols in the range of the mapping, thus we would require the same number of bits to store that information. Hence the average length of compressed data should be greater than or equal to the original average length.

Idea 2: Another idea that pops up in my head is that if the distribution of symbols is uniform, why not create a new symbol set S' that is essentially a set of all the bi-grams of symbols. But, this distribution also needs to be uniform as the probability of a pair of symbols is same for all the possible cases.

These ideas force me to conclude that it is not possible to compress data if the distribution is uniform.

Please correct me if I am wrong at any point, also I would like to know your views on the problem.

Edit: If you can also provide a formal proof or a book for reference then it would be great.

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I'm reading about this in one of the articles that the teacher gave me

Suppose that $({S}_{1},{\mu}_{1})$ and $({S}_{2},{\mu}_{2})$ are two $\sigma $-finite measure spaces and $F:{S}_{1}\times {S}_{2}\to \mathbb{R}$ is measurable. Then

${[{\int}_{{S}_{2}}{|{\int}_{{S}_{1}}F(x,y){\mu}_{1}(\mathrm{d}x)|}^{p}{\mu}_{2}(\mathrm{d}y)]}^{\frac{1}{p}}\le {\int}_{{S}_{1}}{({\int}_{{S}_{2}}|F(x,y){|}^{p}{\mu}_{2}(\mathrm{d}y))}^{\frac{1}{p}}{\mu}_{1}(\mathrm{d}x)$

with obvious modifications in the case $p=\mathrm{\infty}$. If $p>1$, and both sides are finite, then equality holds only if $|F(x,y)|=\phi (x)\psi (y)$ a.e. for some non-negative measurable functions $\phi $ and $\psi $.

If $F$ is non-negative, then the function

$f:y\mapsto {\int}_{{S}_{1}}F(x,y){\mu}_{1}(\mathrm{d}x)$

and thus $|f{|}^{p}$ are measurable. For the LHS to be well-defined, the function $|f{|}^{p}$ should be measurable. In our case, $F$ is not assumed to be non-negative. Could you elaborate on how to prove it?

Suppose that $({S}_{1},{\mu}_{1})$ and $({S}_{2},{\mu}_{2})$ are two $\sigma $-finite measure spaces and $F:{S}_{1}\times {S}_{2}\to \mathbb{R}$ is measurable. Then

${[{\int}_{{S}_{2}}{|{\int}_{{S}_{1}}F(x,y){\mu}_{1}(\mathrm{d}x)|}^{p}{\mu}_{2}(\mathrm{d}y)]}^{\frac{1}{p}}\le {\int}_{{S}_{1}}{({\int}_{{S}_{2}}|F(x,y){|}^{p}{\mu}_{2}(\mathrm{d}y))}^{\frac{1}{p}}{\mu}_{1}(\mathrm{d}x)$

with obvious modifications in the case $p=\mathrm{\infty}$. If $p>1$, and both sides are finite, then equality holds only if $|F(x,y)|=\phi (x)\psi (y)$ a.e. for some non-negative measurable functions $\phi $ and $\psi $.

If $F$ is non-negative, then the function

$f:y\mapsto {\int}_{{S}_{1}}F(x,y){\mu}_{1}(\mathrm{d}x)$

and thus $|f{|}^{p}$ are measurable. For the LHS to be well-defined, the function $|f{|}^{p}$ should be measurable. In our case, $F$ is not assumed to be non-negative. Could you elaborate on how to prove it?

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$\{573,\text{}951,\text{}825,\text{}639\}$