Recent questions in Composite functions

PrecalculusAnswered question

LoomiTymnk63x 2023-03-29

Whether f is a function from Z to R if

a) $f\left(n\right)=\pm n$.

b) $f\left(n\right)=\sqrt{{n}^{2}+1}$.

c) $f\left(n\right)=\frac{1}{{n}^{2}-4}$.

PrecalculusAnswered question

Preston Walker 2023-02-13

Is 93 a prime or composite number? A)Prime number; B)Composite number; C)Cannot be determined

PrecalculusAnswered question

Mara Boyd 2023-01-05

If f(x)=x^2−1/x and g(x)=x+2/x−3, then the domain of f(x)/g(x) is...

PrecalculusAnswered question

Mark Rosales 2022-11-18

How to find domain of complicated composite functions. Find the domain of arccos(${e}^{x}$), are there universal steps I can take to be able to find the domain?

PrecalculusOpen question

2022-11-09

write an equation for a rational function with:

Vertical Asymptotes at x=2 and x=-6

x-intercepts at x=-4 and x=-3

y-intercept at 3

PrecalculusAnswered question

Martin Hart 2022-10-31

Question about calculating the gradient of a composite function

$\mathbf{r}=(x,y)=x\mathbf{i}+y\mathbf{j}$

$\Vert r\Vert =\sqrt{{x}^{2}+{y}^{2}}$

Let's assume that $\mathbf{r}\ne 0$

We now define $f(x,y)={r}^{m}$

What is the right expression for $\mathrm{\nabla}f$?

1. $m{r}^{m-1}\mathbf{r}$

2. $m{r}^{m-2}\mathbf{r}$

3. $m{r}^{m-1}$

4. $m{r}^{0.5m-1}\mathbf{r}$

Reasoning was that it should be number $$1$$:

the derivative of $f$ according to $x$ comes out as $m{r}^{m-1}(\frac{x}{\sqrt{{x}^{2}+{y}^{2}}})$, while the derivative of $f$ according to y comes out as $m{r}^{m-1}(\frac{y}{\sqrt{{x}^{2}+{y}^{2}}})$. $\sqrt{{x}^{2}+{y}^{2}}=\Vert r\Vert =1$, and therefore this is equal to $m{r}^{m-1}\mathbf{r}$.

$\mathbf{r}=(x,y)=x\mathbf{i}+y\mathbf{j}$

$\Vert r\Vert =\sqrt{{x}^{2}+{y}^{2}}$

Let's assume that $\mathbf{r}\ne 0$

We now define $f(x,y)={r}^{m}$

What is the right expression for $\mathrm{\nabla}f$?

1. $m{r}^{m-1}\mathbf{r}$

2. $m{r}^{m-2}\mathbf{r}$

3. $m{r}^{m-1}$

4. $m{r}^{0.5m-1}\mathbf{r}$

Reasoning was that it should be number $$1$$:

the derivative of $f$ according to $x$ comes out as $m{r}^{m-1}(\frac{x}{\sqrt{{x}^{2}+{y}^{2}}})$, while the derivative of $f$ according to y comes out as $m{r}^{m-1}(\frac{y}{\sqrt{{x}^{2}+{y}^{2}}})$. $\sqrt{{x}^{2}+{y}^{2}}=\Vert r\Vert =1$, and therefore this is equal to $m{r}^{m-1}\mathbf{r}$.

PrecalculusAnswered question

adarascarlet80 2022-10-05

Suppose that we have three $\mathbb{Z}\to \mathbb{Z}$ functions such as $f$, $g$ and $h$. How should $f$ and $h$ be so that f∘g∘h can be onto (surjective) given that $g$ is a one to one (injective) function?

PrecalculusAnswered question

trapskrumcu 2022-09-26

Chain rule for the derivative of a composite function

$$y=(\mathrm{sin}x{)}^{\sqrt{x}}.$$

$$y=(\mathrm{sin}x{)}^{\sqrt{x}}.$$

PrecalculusAnswered question

gaby131o 2022-09-26

Show that if $\underset{x\to a}{lim}f(x)=L$, then $\underset{x\to a}{lim}cos(f(x))=cos(L)$.

PrecalculusAnswered question

videosfapaturqz 2022-09-24

Let $f:\mathbb{D}\to \mathbb{D}$ (unit disk) be a holomorphic function with $f(0)=0,|{f}^{\prime}(0)|<1$. For ${f}_{n}=f\circ \cdots \circ f$, show that $\sum _{n=1}^{\mathrm{\infty}}{f}_{n}(z)$ converges uniformly on compact subsets in $\mathbb{D}$.

I tried Schwarz lemma so that $|f(z)|\le |z|$, and I tried to use Weierstrass $$M$$ test, but I don't know how ${f}_{n}$ is bounded. How to solve this problem?

I tried Schwarz lemma so that $|f(z)|\le |z|$, and I tried to use Weierstrass $$M$$ test, but I don't know how ${f}_{n}$ is bounded. How to solve this problem?

PrecalculusAnswered question

malaana5k 2022-09-24

Lets have $y:\mathbb{R}\to {\mathbb{R}}^{2}$ and that $f:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$, and lets assume that $f(y(x))$ is given and that $y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}f(y(t))dt$

I'm a bit confused how there can be a function of $y(t)$ inside of the function definition for $y(x)$.

I took the example that $y(x)=({x}^{2},x)$ and $f(y,z)=(y+z,y-z)$

$$\Rightarrow f(y(x))=f({x}^{2},x)=({x}^{2}+x,{x}^{2}-x)$$

And now if we follow the definition of $y(x)$ we get:

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}f(y(t))dt$$

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}({t}^{2}+t,{t}^{2}-t)dt$$

$$\Rightarrow y(x)=y({x}_{0})+(\frac{1}{3}{t}^{3}+\frac{1}{2}{t}^{2},\frac{1}{3}{t}^{3}-\frac{1}{2}{t}^{2}){{\textstyle |}}_{{x}_{0}}^{x}$$

$$\Rightarrow y(x)=(\frac{1}{3}{x}^{3}+\frac{1}{2}{x}^{2}+{C}_{1},\frac{1}{3}{x}^{3}-\frac{1}{2}{x}^{2}+{C}_{2})$$

Where is mistake?

I'm a bit confused how there can be a function of $y(t)$ inside of the function definition for $y(x)$.

I took the example that $y(x)=({x}^{2},x)$ and $f(y,z)=(y+z,y-z)$

$$\Rightarrow f(y(x))=f({x}^{2},x)=({x}^{2}+x,{x}^{2}-x)$$

And now if we follow the definition of $y(x)$ we get:

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}f(y(t))dt$$

$$y(x)=y({x}_{0})+{\int}_{{x}_{0}}^{x}({t}^{2}+t,{t}^{2}-t)dt$$

$$\Rightarrow y(x)=y({x}_{0})+(\frac{1}{3}{t}^{3}+\frac{1}{2}{t}^{2},\frac{1}{3}{t}^{3}-\frac{1}{2}{t}^{2}){{\textstyle |}}_{{x}_{0}}^{x}$$

$$\Rightarrow y(x)=(\frac{1}{3}{x}^{3}+\frac{1}{2}{x}^{2}+{C}_{1},\frac{1}{3}{x}^{3}-\frac{1}{2}{x}^{2}+{C}_{2})$$

Where is mistake?

PrecalculusAnswered question

Camila Brandt 2022-09-23

Find the indefinite integral of $x\times (5x-1{)}^{19}$ by substitution

My try:

$u=5x-1$, so $\frac{du}{dx}=5$, thus $dx=\frac{du}{5}$

How to cancel out the $x$ in front?

My try:

$u=5x-1$, so $\frac{du}{dx}=5$, thus $dx=\frac{du}{5}$

How to cancel out the $x$ in front?

PrecalculusAnswered question

unjulpild9b 2022-09-20

Chain rule of partial derivatives for composite functions.

Function of the form

$$f({x}^{2}+{y}^{2})$$

How do I find the partial derivatives

$$\frac{\mathrm{\partial}f}{\mathrm{\partial}y},\frac{\mathrm{\partial}f}{\mathrm{\partial}x}$$

How $f({x}^{2}+{y}^{2})$ behaves. Assuming it should of the form

$$g(x,y)\cdot 2y\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}h(x,y)\cdot 2x$$

Function of the form

$$f({x}^{2}+{y}^{2})$$

How do I find the partial derivatives

$$\frac{\mathrm{\partial}f}{\mathrm{\partial}y},\frac{\mathrm{\partial}f}{\mathrm{\partial}x}$$

How $f({x}^{2}+{y}^{2})$ behaves. Assuming it should of the form

$$g(x,y)\cdot 2y\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}h(x,y)\cdot 2x$$

PrecalculusAnswered question

Liam Potter 2022-09-20

Two continuous functions $f(x)$ and $g(x)$, is it possible that I expand $f(g(x))$ at $g(0)$ using a series of $g(x)$?

For example,

$$f(g(x))=f(g(0))+{f}^{\prime}(g(0))g(x)+\frac{{f}^{\u2033}(g(0))}{2}{g}^{2}(x)+\cdots $$

In my case, $g(x)={e}^{-{x}^{2}}(0\le x\le 1)$.

For example,

$$f(g(x))=f(g(0))+{f}^{\prime}(g(0))g(x)+\frac{{f}^{\u2033}(g(0))}{2}{g}^{2}(x)+\cdots $$

In my case, $g(x)={e}^{-{x}^{2}}(0\le x\le 1)$.

PrecalculusAnswered question

Aidyn Meza 2022-09-20

Trying to calculate the value of $$\frac{{\pi}^{4}}{90}$$. Although I know the exact value (which I found on google to be $$\frac{{\pi}^{4}}{90}$$) but I wanted to derive it by myself. While doing so, I arrived at this rather peculiar expression: $C=\frac{7{\u213c}^{4}}{720}-\frac{1}{2}-\frac{P}{2}$

where $$C$$ is the value of the composite zeta function at $$2$$ and $$P$$ is the prime zeta function at $$2$$. My question is this. What will be the value of $$C$$?

where $$C$$ is the value of the composite zeta function at $$2$$ and $$P$$ is the prime zeta function at $$2$$. My question is this. What will be the value of $$C$$?

PrecalculusAnswered question

vballa15ei 2022-09-14

The question is:

$f(x)={\displaystyle \frac{x}{x-1}}$

$g(x)={\displaystyle \frac{1}{x}}$

$h(x)={x}^{2}-1$

Find $f\circ g\circ h$ and state its domain.

The answer the textbook states is that the domain is all real values of $x$, except $\pm 1$ and $\pm \sqrt{2}$.

However surely the domain excludes $0$ as well, since $g(0)$ is undefined.

$f(x)={\displaystyle \frac{x}{x-1}}$

$g(x)={\displaystyle \frac{1}{x}}$

$h(x)={x}^{2}-1$

Find $f\circ g\circ h$ and state its domain.

The answer the textbook states is that the domain is all real values of $x$, except $\pm 1$ and $\pm \sqrt{2}$.

However surely the domain excludes $0$ as well, since $g(0)$ is undefined.

PrecalculusAnswered question

Modelfino0g 2022-09-14

Thorem: If $f(x)$ is continuous at $L$ and $\underset{x\to a}{lim}g(x)=L$, then $\underset{x\to a}{lim}f(g(x)=f(\underset{x\to a}{lim}g(x))=f(L)$.

Proof: Assume $f(x)$ is continuous at a point $L$, and that $\underset{x\to a}{lim}g(x)=L$.

$\mathrm{\forall}\u03f5>0,\mathrm{\exists}\delta >0:[|x-L|<\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\textstyle |}f(x)-f(L){\textstyle |}<\epsilon ]$.

And $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|g(x)-L<\delta ]$.

So, $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(g(x))-f(L)|<\u03f5]$.

$\underset{x\to a}{lim}g(x)=L$ so $f(\underset{x\to a}{lim}g(x))=f(L)$.

Proof: Assume $f(x)$ is continuous at a point $L$, and that $\underset{x\to a}{lim}g(x)=L$.

$\mathrm{\forall}\u03f5>0,\mathrm{\exists}\delta >0:[|x-L|<\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\textstyle |}f(x)-f(L){\textstyle |}<\epsilon ]$.

And $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|g(x)-L<\delta ]$.

So, $\mathrm{\forall}\delta >0,\mathrm{\exists}{\delta}^{\prime}>0:[|x-a|<{\delta}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(g(x))-f(L)|<\u03f5]$.

$\underset{x\to a}{lim}g(x)=L$ so $f(\underset{x\to a}{lim}g(x))=f(L)$.

PrecalculusAnswered question

spremani0r 2022-09-13

Need to find $f(x{)}^{\prime}$ while $f(x)=ln(x+\sqrt{{a}^{2}+{x}^{2}})$

I have $f(x{)}^{\prime}=\frac{1}{(x+\sqrt{{a}^{2}+{x}^{2}})}\cdot (1+\frac{2x}{2\sqrt{{a}^{2}+{x}^{2}}})$, but can't simplify.

I want get $\frac{1}{\sqrt{{a}^{2}+{x}^{2}}}$

I have $f(x{)}^{\prime}=\frac{1}{(x+\sqrt{{a}^{2}+{x}^{2}})}\cdot (1+\frac{2x}{2\sqrt{{a}^{2}+{x}^{2}}})$, but can't simplify.

I want get $\frac{1}{\sqrt{{a}^{2}+{x}^{2}}}$

PrecalculusAnswered question

Spactapsula2l 2022-09-12

Derivative of this trig function is:

$$\begin{array}{rl}\frac{d}{dx}[\mathrm{sec}\left(\frac{x}{12}\right)]\text{}& =\mathrm{sec}\left(\frac{x}{12}\right)\ast \mathrm{tan}\left(\frac{x}{12}\right)\ast \frac{d}{dx}\left(\frac{x}{12}\right)\\ & =\mathrm{sec}\left(\frac{x}{12}\right)\ast \mathrm{tan}\left(\frac{x}{12}\right)\ast \frac{1}{12}\end{array}$$

If chain rule is not applied to this function like this because the function is "composite" which is why it should be done as the first variant, then how was chain rule altered for this function in the first variant?

$$\begin{array}{rl}\frac{d}{dx}[\mathrm{sec}\left(\frac{x}{12}\right)]\text{}& =\mathrm{sec}\left(\frac{x}{12}\right)\ast \mathrm{tan}\left(\frac{x}{12}\right)\ast \frac{d}{dx}\left(\frac{x}{12}\right)\\ & =\mathrm{sec}\left(\frac{x}{12}\right)\ast \mathrm{tan}\left(\frac{x}{12}\right)\ast \frac{1}{12}\end{array}$$

If chain rule is not applied to this function like this because the function is "composite" which is why it should be done as the first variant, then how was chain rule altered for this function in the first variant?

PrecalculusAnswered question

nar6jetaime86 2022-09-12

Let $A$, $B$ & $C$ sets, and left $f:A\to B$ and $g:B\to C$ be functions. Suppose that $f$ and $g$ have inverses. Prove that $g\circ f$ has an inverse, and that $(g\circ f{)}^{-1}={f}^{-1}\circ {g}^{-1}$.

Assuming that f and g have reverse, ${f}^{-1}=h$ and ${g}^{-1}=s$ with $h:B\to A$, $s:C\to B$.

from that above i infer that the inverse of $(g\circ f)$ is $(s\circ g):C\to A$ that is ${g}^{-1}\circ {f}^{-1}=(g\circ f{)}^{-1}$; Hence for proof of $(g\circ f{)}^{-1}={f}^{-1}\circ {g}^{-1}$, proceed as before, only swapping functions, right?

Assuming that f and g have reverse, ${f}^{-1}=h$ and ${g}^{-1}=s$ with $h:B\to A$, $s:C\to B$.

from that above i infer that the inverse of $(g\circ f)$ is $(s\circ g):C\to A$ that is ${g}^{-1}\circ {f}^{-1}=(g\circ f{)}^{-1}$; Hence for proof of $(g\circ f{)}^{-1}={f}^{-1}\circ {g}^{-1}$, proceed as before, only swapping functions, right?

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