Invertibility of f(x) and its Derivative g'(y)

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In summary, we are given a function f(x) = 1 + \int_0^xe^{t^2} dt where x E \Re and the interval for this function goes from (0,x). We are asked to prove that the range of f is \Re and that f: \Re \rightarrow\Re is invertible. We are also asked to find the inverse of f, denoted by g, and show that g is differentiable and satisfies the equation g'(y) = e^-(g(y))^2 for all y E \Re. Finally, we need to determine the values of g(1), g'(1), and g"(1). To prove that the range of f is \Re
  • #1
tomboi03
77
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For x E [tex]\Re[/tex], let
f(x) = 1 + [tex]\int[/tex]et2 dt
(the interval for this function goes from (0,x) i just didn't know how to put it on the integral.)

i. Prove that the range of f is [tex]\Re[/tex] (i.e. prove that for every y E [tex]\Re[/tex] there is an x E [tex]\Re[/tex] such that f(x)=y)

ii. Prove that f: [tex]\Re[/tex] [tex]\rightarrow[/tex][tex]\Re[/tex] is invertible

iii. Denote the inverse of f by g. Argue that g is differentiable and show that g satisfies the equation
g'(y) = e-(g(y))2
for all y E [tex]\Re[/tex]. Show that g is differentiable twice.

iv. Determine g(1), g'(1), g"(1).

okay, so for i, i have no idea
ii, how can you prove that a function is invertible
iii, i have no idea
iv, i just find the first derivative and the 2nd derivative and find the values of all of that.

Please help me out,
Thank You
 
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  • #2
tomboi03 said:
For x E [tex]\Re[/tex], let
f(x) = 1 + [tex]\int[/tex]et2 dt
(the interval for this function goes from (0,x) i just didn't know how to put it on the integral.)

i. Prove that the range of f is [tex]\Re[/tex] (i.e. prove that for every y E [tex]\Re[/tex] there is an x E [tex]\Re[/tex] such that f(x)=y)

Hi tomboi03! :smile:

(have an ε :wink:)

Do you mean [tex]f(x)\ =\ 1\ +\ \int_0^xe^{t^2} dt[/tex] ?

Hint: what are f(-∞) and f(∞)? :wink:
 
  • #3


nope, it's still et2

hahahaha :D hehehe
Thanks for helping me! i really appreciate it! :D hehehe
 
  • #4


Can someone help me with this.. the previous person didn't help me much... thank you
 
  • #5


tomboi03 said:
For x E [tex]\Re[/tex], let
f(x) = 1 + [tex]\int[/tex]et2 dt
(the interval for this function goes from (0,x) i just didn't know how to put it on the integral.)

i. Prove that the range of f is [tex]\Re[/tex] (i.e. prove that for every y E [tex]\Re[/tex] there is an x E [tex]\Re[/tex] such that f(x)=y)

ii. Prove that f: [tex]\Re[/tex] [tex]\rightarrow[/tex][tex]\Re[/tex] is invertible

iii. Denote the inverse of f by g. Argue that g is differentiable and show that g satisfies the equation
g'(y) = e-(g(y))2
for all y E [tex]\Re[/tex]. Show that g is differentiable twice.

iv. Determine g(1), g'(1), g"(1).

okay, so for i, i have no idea
ii, how can you prove that a function is invertible
iii, i have no idea
iv, i just find the first derivative and the 2nd derivative and find the values of all of that.

Please help me out,
Thank You

OK, so apparently [tex]f(x) = \int_0^x et^2 dt[/tex]
Have you gone so far as to evaluate this integral?
 
  • #6


actually it's e^(t^2)...
my professor has changed it.
 
  • #7


Yes, that makes more sense. Which of the questions are you still having problems with?
 
  • #8
tomboi03 said:
actually it's e^(t^2)...
my professor has changed it.

mmm … that's what i said! :rolleyes:
tiny-tim said:
Do you mean [tex]f(x)\ =\ 1\ +\ \int_0^xe^{t^2} dt[/tex] ?

Hint: what are f(-∞) and f(∞)? :wink:

ok … for
i. Prove that the range of f is R (i.e. prove that for every y E there is an x E such that f(x)=y)​
… find f(-∞) and f(∞) and describe how it goes from one to the other. :smile:
 
  • #9


i'm not sure if i understand how to solve the integral...

and i was wondering... does invertible mean show that it is bijective?

I'm confused.
 
  • #10
tomboi03 said:
i'm not sure if i understand how to solve the integral...

Hi tomboi03! :smile:

You don't need to "solve" it … just look at it :smile: … what is its value at ∞?
and i was wondering... does invertible mean show that it is bijective?

Invertible means it has a unique inverse, so if it's onto and invertible, then yes, it's bijective. :wink:
 
  • #11


is it infinity? I'm not sure how i can just loook at it... :'(
 
  • #12
tomboi03 said:
is it infinity? I'm not sure how i can just loook at it... :'(

:biggrin:

well, if you draw it, the integral is the area under it (to the right of the y-axis) …

isn't that obviously ∞?

('cos it keeps going up! :rolleyes:)

ok, now what's f(-∞)? :smile:
 
  • #13


isn't it negative infiinity?
 
  • #14


so how do you prove that this is invertible?
 
  • #15
tomboi03 said:
isn't it negative infiinity?

Yup! :biggrin:
tomboi03 said:
so how do you prove that this is invertible?

Well, it goes from -∞ to ∞, so all you have to prove is that it doesn't go through any level twice. :wink:
 

What is an invertible math problem?

An invertible math problem is a mathematical equation or expression that can be solved both forwards and backwards. This means that given the solution, you can find the original equation, and vice versa.

What makes a math problem invertible?

A math problem is invertible if it has a one-to-one relationship between the input and output. This means that each input has a unique output, and each output has a unique input.

Why are invertible math problems important?

Invertible math problems are important because they allow for efficient problem solving and can provide a deeper understanding of mathematical concepts. They are also used in various fields such as computer science, engineering, and physics.

What are some examples of invertible math problems?

Some examples of invertible math problems are linear equations, quadratic equations, and logarithmic equations. These equations can be solved in both directions, and the solutions can be verified by substituting them back into the original equation.

Can all math problems be inverted?

No, not all math problems can be inverted. Invertibility depends on the type of equation and the relationship between the input and output. Some equations, such as those with multiple solutions or no solutions, cannot be inverted.

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