If the marginal income function is known, the income function can be obtained A detailed explanation of the question is helpful for the respondent to give an accurate answer

The profit function can be obtained by integrating the marginal profit function
For example, the marginal revenue function of a manufacturer is Mr = 10-2q
The return function is r = 10r-q ^ 2

High number, calculus. Write marginal revenue function, Mr = R '(q) and R' (P) are OK?

Indicates always write R (P)

How to find the inverse demand function from the marginal revenue function
For example: the marginal revenue function Mr = 120-6q, we can get the inverse demand function p = 120-3q?

Let Q + B be a constant
R=PQ=(aQ+b)Q
MR is the derivative of R = 2aq + B
So 2A = - 6, B = 120
Substituting P = AQ + B
Get the answer

Higher number function
It is known that f (x) is a linear function, and for any t ∈ R, there is always 3f (T + 1) - 2F (t-1) = 2T + 17
Finding the expression of F (x)

F (x) is a linear function, Let f (x) = ax + B
3f(t+1)-2f(t-1)=3[a(t+1)+b]-2[a(t-1)+b]=at+5a+b=2t+17
a=2 b=7
f(x)=2x+7

Advanced function problems
Find the limit of (2 ^ n + 3 ^ n) / {(2 ^ n + 1) + (3 ^ n + 1)} when x tends to infinity

The numerator and denominator are divided by (3 ^ n + 1) again: 1 / 3

Functions in higher numbers,
x＋1 ＜0
Let f (x)={
If X-1 ≤ 0, the X of limf (x) tends to be the same as that of 0 ()
(A) 1; (b) - 1; (c) 1 and - 1; (d) do not exist
It's a piecewise function. I think it's x + 1 < 0, and the left limit is 1,
There is no right limit when X-1 ≤ 0, because I think there should be right limit only when X-1 ≤ 0 is greater than zero?

Did you write the title wrong?
Is this a piecewise function
The definition field of the first function should be x > 0! Or change the following!
If so, then when x = 0, you need to consider its left and right limits
Obviously, the left limit is substituted into the first function, x + 1 is 0 + 1 = 1
The right limit is substituted into the second function, X-1 is 0-1 = - 1
The left and right limits are not equal! Then they don't exist!
Got it! Is the title right? Let's see if it is!
If you have any questions, leave me a message!

███████████ (novice) ███
The range of y = f (x) is [1,3], then the range of F (x) = 1-2f (x + 3) is
What is the answer? I know the answer;
They have the same range of values, but I think their corresponding relations are different. Why do they have the same range of values? Help! Help!

The same value range means the same value range of Y and different function formula means the same value of Y and different value of X
Take the cube of two functions y = x and y = x as an example
The range of values of these two functions, that is, the range of y value, is from negative infinity to positive infinity, and all real numbers can be obtained. Therefore, the range of values of the two functions is the same. But when it comes to a specific y value, for example, y = 27, x = 27 is required for y = x, and x = 3 is required for y = x cube. However, both y = x and y = x cube have the opportunity to let y = 27
But the ranges of y = x and y = x & # 178; are different. For y = x, y can take the value of - 3; but for y = x & # 178; function, y can't be equal to - 3. That's why the ranges of Y values are different
So different functions, the range of Y values can be exactly the same, because it does not mean that the same x gets the same y

There is a problem f (x) in a certain interval for piecewise function and continuous expression, I will not write
If the breakpoint is x = 0, then we can find the integral expression of each part (except x = 0)
When f (x) is at x0, two integral expressions are equal at x = 0 (limit)
Why
If the solution seems to say that f (x) is continuous, then we can deduce the continuity of his original function

If the integrand function is defined at x = 0, then the original function must be differentiable at this point, because the derivative of the original function is equal to the integrand function, and the same is true at x = 0, which indicates that the original function must be continuous at x = 0

A super simple function of higher numbers
Given that the domain of the function y = f (x) is [0,1], then the domain of F (x squared) is [- 1,1]. How can this be calculated

The domain of F (x) is [0,1], then the domain of y = f (X & sup2;) is [- 1,1],
Let x = A & sup2;, then a & sup2; ∈ [0,1]
0＜a²＜1,
∴a∈[-1,1]
The domain of definition of F (a) is [- 1,1],
Let a = x & sup2;, so the domain of F (X & sup2;) is [- 1,1]
Maybe you don't understand how the last step came out,
This is the definition of function domain. Here, f (x) is different from X in F (X & sup2;). What we don't understand is that we often confuse two X, so it's convenient for you to set an X as a. let's understand it slowly... If you don't understand, let's take a look at how to define domain in books
I've read the one on the fourth floor and the one on the fourth floor. It's easier to understand it. However, the problem is "the domain of definition of known function y = f (x) is [0,1], and the domain of definition of finding f (x squared) is [- 1,1]". I don't think it's right to deduce

Ask a simple function problem
How to find the interval with y = t + T ^ - 2 value greater than 0 and the interval with y = t + T ^ - 2 value less than 0?

It is easy to know that the domain of definition does not include 0. Let's sort out the formula a little bit, y = (T ^ 3 + 1) / T ^ 2, if the denominator is greater than 0, then the interval greater than 0: T ^ 3 + 1 > 0, so - 1

If the marginal income function is known, the income function can be obtained A detailed explanation of the question is helpful for the respondent to give an accurate answer