If the function y = f (x) has an inverse function y = F-1 (x), find the inverse function of y = f (x-1)? Thank you!

If the function y = f (x) has an inverse function y = F-1 (x), find the inverse function of y = f (x-1)? Thank you!

Drawing
Functions and inverse functions are symmetric about y = X
So the original function is to shift the image to the right one unit along the X axis
So the inverse function image is translated up one unit along the Y axis
So Y-1 = F-1 (x)
y=f-1(x)+1

The definition domain of F (x) = x ^ 2 + 4 (a + 1) x-3 is [- 2,2]. This function has an inverse function y = f ^ - 1 (x). Find the value range of A

The definition domain of F (x) = x ^ 2 + 4 (a + 1) x-3 is [- 2,2], and there is an inverse function y = f ^ - 1 (x), which is monotonic in the definition field. For a quadratic function, as long as the definition field does not include the symmetry axis, 2 (a + 1) > = 2 or 2 (a + 1) = 0 or a

If y = f (x) has an inverse function, then y = inverse function + 1 image must cross the fixed point?

Primitive function and inverse function are symmetric about Y-X = 0
The symmetric point of (0,1) with respect to y = x is (1,0)
So the inverse function is over (1,0)
G (x) over (1,0)
g(1)=0
y=Q(x)=g(x)+1
Q(1)=1
So pass (1,1)

Is the inverse function of the function y = 2x + 1? I overslept today, who can teach me the inverse function and teach me this problem! Now I'm upset by your anger. I said that I can't inverse function, I don't know the concept. You just give a string and you can't write it clearly. You think I can understand it. If you want to be angry, I won't answer it, OK?

The inverse function of y = 2x + 1 is y = 0.5x-0.5. Generally, let the value range of the function y = f (x) (x ∈ a) be c. according to the relationship between X and Y in this function, X is expressed by Y, and x = f (y) is obtained

Find the inverse function of y = x ^ 2 + 2x-1 (x > 0)

y=(x+1)²-1-1
(x+1)²=y+2
X = radical y + 2-1
XY interchange
Y = radical y + 2) - 1

y=(2x-1)^2(x

y=(x+1)²-1-1
(x+1)²=y+2
X = radical y + 2-1
XY interchange
Y = radical y + 2) - 1
I hope it will help you

Find y = (x ^ 2) + 2x (x ≥ 0) y = - x + 2x (x)

y=x²+2x
=(x+1)²-1
x>=0 ;y>=0
y+1=(x+1)²
√(y+1)=x+1
x=√(y+1)-1
The inverse function is y = √ (x + 1) - 1
y=-x²+2x
=-(x²-2x+1)+1
=-(x-1)²+1
X

Given the function y = 2x + 1 (- 1 ≤ x ≤ 2), then the inverse function of the function is Given the function y = 2x + 1 (- 1 ≤ x ≤ 2), then the inverse function of the function is

y=2x+1
2x=y-1
x=(y-1)/2
So the inverse function of this function is y = (x-1) / 2
Because the definition domain of inverse function is the value range of original function
-1≤x≤2
-2≤2x≤4
-1≤2x+1≤5
So the definition domain of inverse function is [- 1,5]
The inverse function of the function is y = (x-1) / 2 (- 1 ≤ x ≤ 5)

Inverse function of y = x Λ 2 + 1 / X Λ 2 [x greater than or equal to - 1 less than 0]

When x

Find the inverse function of y = (5 ^ x + 1) / (2 + 5x),

Let a = 5 ^ X
y=(a+1)/(2+a)
2y+ay=a+1
ay-a=1-2y
a(y-1)=1-2y
So a = 5 ^ x = (1-2y) / (Y-1)
x=log5[(1-2y)/(y-1)]
So the inverse function is y = log5 [(1-2x) / (x-1)]