Mastering Composite Functions: Examples, Problems, and Solutions

Whether f is a function from Z to R if
a) $f (n) = \pm n$ .
b) $f (n) = \sqrt{n^{2} + 1}$ .
c) $f (n) = \frac{1}{n^{2} - 4}$ .
?

Preston Walker 2023-02-13

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

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...

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?

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

Martin Hart 2022-10-31

Question about calculating the gradient of a composite function
$r = (x, y) = x i + y j$
$‖ r ‖ = \sqrt{x^{2} + y^{2}}$
Let's assume that $r \neq 0$
We now define $f (x, y) = r^{m}$
What is the right expression for $\nabla f$ ?
1. $m r^{m - 1} r$
2. $m r^{m - 2} r$
3. $m r^{m - 1}$
4. $m r^{0.5 m - 1} r$
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 $m r^{m - 1} (\frac{y}{\sqrt{x^{2} + y^{2}}})$ . $\sqrt{x^{2} + y^{2}} = ‖ r ‖ = 1$ , and therefore this is equal to $m r^{m - 1} r$ .

adarascarlet80 2022-10-05

Suppose that we have three $Z \to Z$ 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?

trapskrumcu 2022-09-26

Chain rule for the derivative of a composite function
$y = (\sin x)^{\sqrt{x}} .$

gaby131o 2022-09-26

Show that if $lim_{x \to a} f (x) = L$ , then $lim_{x \to a} c o s (f (x)) = c o s (L)$ .

videosfapaturqz 2022-09-24

Let $f : D \to D$ (unit disk) be a holomorphic function with $f (0) = 0, | f^{'} (0) | < 1$ . For $f_{n} = f \circ \dots \circ f$ , show that $\sum_{n = 1}^{\infty} f_{n} (z)$ converges uniformly on compact subsets in $D$ .
I tried Schwarz lemma so that $| f (z) | \leq | z |$ , and I tried to use Weierstrass $M$ test, but I don't know how $f_{n}$ is bounded. How to solve this problem?

malaana5k 2022-09-24

Lets have $y : R \to R^{2}$ and that $f : R^{2} \to R^{2}$ , and lets assume that $f (y (x))$ is given and that $y (x) = y (x_{0}) + \int_{x_{0}}^{x} f (y (t)) d t$
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:
$y (x) = y (x_{0}) + \int_{x_{0}}^{x} f (y (t)) d t$
$y (x) = y (x_{0}) + \int_{x_{0}}^{x} (t^{2} + t, t^{2} - t) d t$
$\Rightarrow y (x) = y (x_{0}) + (\frac{1}{3} t^{3} + \frac{1}{2} t^{2}, \frac{1}{3} t^{3} - \frac{1}{2} t^{2}) |_{x_{0}}^{x}$
$\Rightarrow y (x) = (\frac{1}{3} x^{3} + \frac{1}{2} x^{2} + C_{1}, \frac{1}{3} x^{3} - \frac{1}{2} x^{2} + C_{2})$
Where is mistake?

Camila Brandt 2022-09-23

Find the indefinite integral of $x \times (5 x - 1)^{19}$ by substitution
My try:
$u = 5 x - 1$ , so $\frac{d u}{d x} = 5$ , thus $d x = \frac{d u}{5}$
How to cancel out the $x$ in front?

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
$\frac{\partial f}{\partial y}, \frac{\partial f}{\partial x}$
How $f (x^{2} + y^{2})$ behaves. Assuming it should of the form
$g (x, y) \cdot 2 y or h (x, y) \cdot 2 x$

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,
$f (g (x)) = f (g (0)) + f^{'} (g (0)) g (x) + \frac{f^{″} (g (0))}{2} g^{2} (x) + \dots$
In my case, $g (x) = e^{- x^{2}} (0 \leq x \leq 1)$ .

Aidyn Meza 2022-09-20

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

vballa15ei 2022-09-14

The question is:
$f (x) = \frac{x}{x - 1}$
$g (x) = \frac{1}{x}$
$h (x) = x^{2} - 1$
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.

Modelfino0g 2022-09-14

Thorem: If $f (x)$ is continuous at $L$ and $lim_{x \to a} g (x) = L$ , then $lim_{x \to a} f (g (x) = f (lim_{x \to a} g (x)) = f (L)$ .
Proof: Assume $f (x)$ is continuous at a point $L$ , and that $lim_{x \to a} g (x) = L$ .
$\forall ϵ > 0, \exists δ > 0 : [| x - L | < δ ⟹ | f (x) - f (L) | < ε]$ .
And $\forall δ > 0, \exists δ^{'} > 0 : [| x - a | < δ^{'} ⟹ | g (x) - L < δ]$ .
So, $\forall δ > 0, \exists δ^{'} > 0 : [| x - a | < δ^{'} ⟹ | f (g (x)) - f (L) | < ϵ]$ .
$lim_{x \to a} g (x) = L$ so $f (lim_{x \to a} g (x)) = f (L)$ .

spremani0r 2022-09-13

Need to find $f (x)^{'}$ while $f (x) = l n (x + \sqrt{a^{2} + x^{2}})$
I have $f (x)^{'} = \frac{1}{(x + \sqrt{a^{2} + x^{2}})} \cdot (1 + \frac{2 x}{2 \sqrt{a^{2} + x^{2}}})$ , but can't simplify.
I want get $\frac{1}{\sqrt{a^{2} + x^{2}}}$

Spactapsula2l 2022-09-12

Derivative of this trig function is:
$\begin{aligned} \frac{d}{d x} [\sec (\frac{x}{12})] & = \sec (\frac{x}{12}) * \tan (\frac{x}{12}) * \frac{d}{d x} (\frac{x}{12}) \\ = \sec (\frac{x}{12}) * \tan (\frac{x}{12}) * \frac{1}{12} \end{aligned}$
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?

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?

The composite functions examples and solutions are vital for the correct understanding of this aspect of Precalculus. Dealing with Composite Functions for your Engineering task will seem much easier as you look through the list of helpful questions and answers.

If something that you need looks similar to what we have included below, start with the analysis of composite aspect by exploring our examples. It will help you receive the best kind of help because the solutions below have the best examples and use of integral calculus in real life.

Precalculus

Practice Exercises for Composite Functions