Practice Exercises for Composite Functions

Whether f is a function from Z to R if
a) f ( n ) = ± n .
b) f ( n ) = n 2 + 1 .
c) f ( n ) = 1 n 2 − 4 .

Preston Walker 2023-02-13

Mara Boyd 2023-01-05

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 ‖ = x 2 + y 2
Let's assume that r ≠ 0
We now define f ( x , y ) = r m
What is the right expression for ∇ 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 ( x x 2 + y 2 ) , while the derivative of f according to y comes out as m r m − 1 ( y x 2 + y 2 ) . 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 → 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

gaby131o 2022-09-26

videosfapaturqz 2022-09-24

Let f : D → D (unit disk) be a holomorphic function with f ( 0 ) = 0 , | f ′ ( 0 ) | < 1 . For f n = f ∘ ⋯ ∘ f , show that ∑ n = 1 ∞ f n ( z ) converges uniformly on compact subsets in D .
I tried Schwarz lemma so that | f ( z ) | ≤ | 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 → R 2 and that f : R 2 → R 2 , and lets assume that f ( y ( x ) ) is given and that y ( x ) = y ( x 0 ) + ∫ 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 )
⇒ 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 ) + ∫ x 0 x f ( y ( t ) ) d t
y ( x ) = y ( x 0 ) + ∫ x 0 x ( t 2 + t , t 2 − t ) d t
⇒ y ( x ) = y ( x 0 ) + ( 1 3 t 3 + 1 2 t 2 , 1 3 t 3 − 1 2 t 2 ) | x 0 x
⇒ y ( x ) = ( 1 3 x 3 + 1 2 x 2 + C 1 , 1 3 x 3 − 1 2 x 2 + C 2 )
Where is mistake?

Camila Brandt 2022-09-23

Find the indefinite integral of x × ( 5 x − 1 ) 19 by substitution
My try:
u = 5 x − 1 , so d u d x = 5 , thus d x = 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
∂ f ∂ y , ∂ f ∂ x
How f ( x 2 + y 2 ) behaves. Assuming it should of the form
g ( x , y ) ⋅ 2 y or h ( x , y ) ⋅ 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 ) + f ″ ( g ( 0 ) ) 2 g 2 ( x ) + ⋯
In my case, g ( x ) = e − x 2 ( 0 ≤ x ≤ 1 ) .

Aidyn Meza 2022-09-20

Trying to calculate the value of π 4 90 . Although I know the exact value (which I found on google to be π 4 90 ) but I wanted to derive it by myself. While doing so, I arrived at this rather peculiar expression: C = 7 ℼ 4 720 − 1 2 − 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 ) = x x − 1
g ( x ) = 1 x
h ( x ) = x 2 − 1
Find f ∘ g ∘ h and state its domain.
The answer the textbook states is that the domain is all real values of x , except ± 1 and ± 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 → a g ( x ) = L , then lim x → a f ( g ( x ) = f ( lim x → a g ( x ) ) = f ( L ) .
Proof: Assume f ( x ) is continuous at a point L , and that lim x → a g ( x ) = L .
∀ ϵ > 0 , ∃ δ > 0 : [ | x − L | < δ ⟹ | f ( x ) − f ( L ) | < ε ] .
And ∀ δ > 0 , ∃ δ ′ > 0 : [ | x − a | < δ ′ ⟹ | g ( x ) − L < δ ] .
So, ∀ δ > 0 , ∃ δ ′ > 0 : [ | x − a | < δ ′ ⟹ | f ( g ( x ) ) − f ( L ) | < ϵ ] .
lim x → a g ( x ) = L so f ( lim x → a g ( x ) ) = f ( L ) .

spremani0r 2022-09-13

Need to find f ( x ) ′ while f ( x ) = l n ( x + a 2 + x 2 )
I have f ( x ) ′ = 1 ( x + a 2 + x 2 ) ⋅ ( 1 + 2 x 2 a 2 + x 2 ) , but can't simplify.
I want get 1 a 2 + x 2

Spactapsula2l 2022-09-12

Derivative of this trig function is:
d d x [ sec ⁡ ( x 12 ) ] = sec ⁡ ( x 12 ) ∗ tan ⁡ ( x 12 ) ∗ d d x ( x 12 ) = sec ⁡ ( x 12 ) ∗ tan ⁡ ( x 12 ) ∗ 1 12
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