# Determine the range of a function using parameter differentiation

• schniefen
In summary: I guess.It's not a bad idea to set up a table of values for this function to get an idea of how it behaves. The function is undefined at x = 0. But as x approaches 0 from the right, the function approaches 0. As x increases, the function increases, reaching a maximum at x = ln(2), the right hand endpoint of the interval. After that the function decreases, reaching 0 at infinity. In summary, the range of f(x) for x>0 is (0, ln(2)).
schniefen
Homework Statement
Determine the range of ##f(x)## for ##x>0## using differentiation under the integral sign.
Relevant Equations
##f(x)=\int_x^{2x} (\frac{e^{-t^2x}}{t}) \ dt ## for ##x>0##.
The strategy here would probably be to find a differential equation that ##f## satisfies, but differentiating with respect to ##x## using Leibniz rule yields

##f'=\int_x^{2x} (-te^{-t^2x}) \ dt + \frac{2e^{-4x^3}-e^{-x^3}}{x}##​

Continuing to differentiate will yield the integral term again (with ##t^3## instead of ##-t##) together with the derivative of the second term above. Is this the right strategy?

As a bonus question. Consider the function ##4t^2e^{-t^2}\cos{(2xt)}##. This is the second derivative w.r.t. ##x## of the function ##e^{-t^2}\cos{(2xt)}##. Now see the attached image. How does one find a function ##g(t)## that satisfies the inequality in the theorem stated in order to justify the differentiation under the integral sign? (##t## ranges from ##0## to ##\infty## and ##x\in\textbf{R}##)

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schniefen said:
Homework Statement: Determine the range of ##f(x)## for ##x>0## using differentiation under the integral sign.
Homework Equations: ##f(x)=\int_x^{2x} (\frac{e^{-t^2x}}{t}) \ dt ## for ##x>0##.

The strategy here would probably be to find a differential equation that ##f## satisfies, but differentiating with respect to ##x## using Leibniz rule yields

##f'=\int_x^{2x} (-te^{-t^2x}) \ dt + \frac{e^{-4x^3}-e^{-x^3}}{x}##
You're making this harder than it needs to be. Finding the derivative is a fairly straightforward application of the Fundamental Theorem of Calculus, one part of which says that
$$\frac d {dx}\int_a^x f(t) dt = f(x)$$
The problem is slightly more complicated in that the integral has functions of x in both of the limits of integration. In this case, you can break up the integral like so:
$$\int_x^{2x} f(t) dt = \int_x^a f(t)dt + \int_a^{2x}f(t)dt$$
schniefen said:
Continuing to differentiate will yield the integral term again (with ##t^3## instead of ##-t##) together with the derivative of the second term above. Is this the right strategy?

As a bonus question. Consider the function ##4t^2e^{-t^2}\cos{(2xt)}##. This is the second derivative w.r.t. ##x## of the function ##e^{-t^2}\cos{(2xt)}##. Now see the attached image. How does one find a function ##g(t)## that satisfies the inequality in the theorem stated in order to justify the differentiation under the integral sign? (##t## ranges from ##0## to ##\infty## and ##x\in\textbf{R}##)

View attachment 251163

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But breaking up the integral still yields

##f'=\int_x^{2x} (-te^{-t^2x}) \ dt + \frac{2e^{-4x^3}-e^{-x^3}}{x}##
, correct?

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There seems to be something strange with having an x as both a limit of integration as well as inside of the expression you are integrating.

schniefen said:
But breaking up the integral still yields

##f'=\int_x^{2x} (-te^{-t^2x}) \ dt + \frac{2e^{-4x^3}-e^{-x^3}}{x}##
, correct?
That's close to but not the same as what I'm getting. In the Liebniz rule, the integrand, f(x, t), is ##\frac{e^{-t^2x}} t##.
What is f(x, b(x))?
What is f(x, a(x))?
What is ##\frac{\partial}{\partial x}f(x, t)##?
In the above, f is different from what you're using to represent the integral.

Mark44 said:
That's close to but not the same as what I'm getting. In the Liebniz rule, the integrand, f(x, t), is ##\frac{e^{-t^2x}} t##.
What is f(x, b(x))?
What is f(x, a(x))?
What is ##\frac{\partial}{\partial x}f(x, t)##?
In the above, f is different from what you're using to represent the integral.
##f(x, b(x))=\frac{e^{-4x^3}}{2x}## (the ##2## in the denominator is eliminated in the Leibniz rule since ##b'(x)=2##
##f(x,a(x))=\frac{e^{-x^3}}{x}##
##\frac{\partial}{\partial x}f(x, t)=(-te^{-t^2x}) ##

schniefen said:
But breaking up the integral still yields

##f'=\int_x^{2x} (-te^{-t^2x}) \ dt + \frac{2e^{-4x^3}-e^{-x^3}}{x}##
, correct?
Wouldn't it be ##f'=\int_x^{2x} (-te^{-t^2x}) \ dt + \frac{e^{-4x^3}-e^{-x^3}}{x}##?
IOW, shouldn't the coefficient of the ##e^{-4x^3}## be 1 instead of 2?

Correct!

Mark44 said:
Wouldn't it be ##f'=\int_x^{2x} (-te^{-t^2x}) \ dt + \frac{e^{-4x^3}-e^{-x^3}}{x}##?
IOW, shouldn't the coefficient of the ##e^{-4x^3}## be 1 instead of 2?
Where would one go from here?

The integral can be evaluated, treating x in the integrand as if it were constant. The substitution u(t) = -t^2x, give du = -2txdt. Find a "t" antiderivative of the integrand, and then evaluate it at the two limits of integration. Once this is done, maybe you can say something about f, due to f' being zero, or positive, or negative.

schniefen
Using u-substitution and adding the other terms to the evaluated integral,

##f'(x)= \frac{3e^{-4x^3}-3e^{-x^3}}{2x}##
The range should be ##(0,\ln{(2)})##. The expression seems kind of challenging to integrate.

## 1. What is parameter differentiation?

Parameter differentiation is a mathematical technique used to find the rate of change in a function with respect to a parameter. It is similar to regular differentiation, but instead of differentiating with respect to a variable, it is done with respect to a parameter.

## 2. How is parameter differentiation used to determine the range of a function?

By using parameter differentiation, we can find the critical points of a function and determine whether they correspond to maximum or minimum values. This information can then be used to determine the range of the function.

## 3. Can any function's range be determined using parameter differentiation?

No, parameter differentiation can only be used to determine the range of continuous functions, where the derivative exists and the function is differentiable with respect to the parameter.

## 4. Are there any other methods to determine the range of a function?

Yes, there are other methods such as finding the zeros of the derivative, using the first or second derivative test, or graphing the function. However, parameter differentiation can be a useful and efficient method in certain cases.

## 5. How does the number of parameters affect the determination of a function's range using parameter differentiation?

The number of parameters in a function does not affect the process of using parameter differentiation to determine the range. The same steps and techniques can be applied regardless of the number of parameters in the function.

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