What is the argument of the function f x 1

We can understand f(x) from the following aspects.

First: an understanding of algebraic formulas. Every algebraic formula is essentially a function.

Like the algebraic formula x2-1, it is a function whose independent variable is x, and for each value of x, x2-1 has a unique value corresponding to it, so the set of all values of x2-1 is the range of the function.

Second, for an abstract function without a concrete analytic formula, it is difficult for us to understand its symbols and meanings because we do not know its specific correspondence laws and its self-variation, definition domain, and value range.

For example, what is the independent variable of f(x+1)? Is its correspondence still f? The independent variable of f(x+1) is x, and its counterpart is not f.

We may assume that if f(x)=x2+1, then f(x+1)=(x+1)2+1, and the algebraic equation f(x+1)2+1 is equal, i.e., the independent variable of (x+1)2+1 is the independent variable of f(x+1). The corresponding rule of (x+1)2+1 is to add 1 to the independent variable and then square it, and then add 1.

For example, are f(x) and f(t) the same function?

All you need to do is give a special description of the function.

Obviously, f(x) and f(t) have the same corresponding rules, if the range of x is the same as the range of t, then f(x) and f(t) are the same functions, otherwise, they are functions with the same corresponding rules but different domains.

For example, if you know that f(x+1)=x +1 and the definition domain of f(x+1) is [0,2], find the f(x) analytic formula and the definition domain.

Let x+1=t, then; x=t-1, then the function of the independent variable f with t is: (i.e., substituting x=t-1 into f(x+1)=x +1).

f(t)=f(x+1)=(t-1)²+1

t²-2t+1+1

t²-2t+2

So, f(t)=t -2t+2, then f(x)=x -2x+2

Or in this way – more intuitive:

Let x=x-1 in f(x+1)=x +1, which is more intuitive, and substitute x=x-1 into f(x+1)=x +1, then:

f(x)=f[(x-1)+1]=(x-1)²+1

x²-2x+1+1

x²-2x+2

So, f(x)=x -2x+2

F(x) and f(t) must have the same range of values as x and t to be the same function, from t=x+1, f(x+1) defines the domain as [0,2], we can know: t [1,3].

f(x)=x -2x+2 is defined in the domain of: x [1,3].

In summary, f(x)=x-2x+2(x [1,3]).

X-1You can set X-1 to U

It becomes f(u).

It's x-1It's a whole, imagine it with a holistic mind.

First question.

f(x+1).

Yes xThis is a composite function of the scrambled.

Because there is no specific function expression.

Therefore, it can only be expressed abstractly by f(x+1).

Second question.

In the symbol f( ), the parentheses indicate that the image of the function f(x) is shifted one unit to the left to form a new celery stove excitation function sock number f(x+1).

The third question.

Just replace x with x+c.

i.e. f(x+c)=k(x+c)+b

Summary. The independent variable of f(x+1) is x

The independent variable of f(t) is t

Is the function f(x+1) argument x or x+1? If we take that one is x 1 equal to t, then what is the independent variable of f (old man t)? For example, f(x)=2 x+1, f(x+1) 2x 3, if x+1 is equal to t, then f(t) is equal to 2t+1.

If the independent variable of x is 2 x Lao Li Sheng4, then I can cite how much of the cherry blossoms, when I think t, it is the same as two x+1, but when I t=x+1, these two are not the same function.

The independent variable of f(x+1) is the independent variable of xf(t).

I'm a little confused below, can you tell me more about it, thank you!

Are f(t)=2 t+1 and fx=2 x+1 the same function?

Yes, it's just that the letters are different.

t=x+1, f(x+1) is equal to 2x 3, isn't He Tuanqing Chanlu and f(x)=2 x1 different functions, and it becomes the same function by changing the hidden number band.

The X in front and the X in the back are not the same, and the range of the two of them is not the same. If the range of the previous t is 1 to 2, then the x in t x+1 is 0 to 1, and the range of x in f(x) 2x+1 is 1 to 2

Is it the same function after replacing it with t?

Yes, after replacing it with t, t and x are in the same range, the range of independent variables is the same, the formula is the same, and the functions are equal.

f(x-1) x-1 is the independent variable x*x is the dependent variable and the slag is round and rolling.

Let t=x-1 then x=t+1

f(t)=(t+1)^2

That is, f(x)=(x+1) 2 (where x is the independent variable, not the x in your question).

Summary. When x=5, the argument in the function f(x+1)=2x-1 should be understood as:

If you want to solve it, you need to follow the steps above.

The variable y constant is 100

Independent variable: x function:

y=10x+100

There is a functional relation of jujube search y=

2x 2-x+2, the constant has 2,2, the variable has x,y, and the independent variable is x; The dependent variable is y; -2x 2-x+2 is a function and y is a function of x.

What does not change is the constant, what can change is the variable, what the stool friend actively changes is the independent variable, and the passive change is the dependent variable; Correspondence is function!