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y=2/3 sinx, -π/2≤x≤π/2

F (x-1) =x ＾ 2-2x+3 (x ≤ 0) for F ＾ -1 (x) and f ＾ -1 (x+1) If f ＾ 1 (x) is read as the inverse of F, then it is the inverse function

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The inverse function is g (x)
y1>y2
x1=f(y1)>x2=f(y2)
Then X1 > x2
g(x1)=y1>g(x2)=y2
So it's an increasing function

How to find the inverse function of y = (X-2) divided by (3x + 4)

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y=(2x+3)/(x-1)
y(x-1)=2x+3
yx-y=2x+3
yx-2x=3+y
(y-2)x=3+y
x=(y+3)/(y-2)
therefore
The inverse function is
y=(x+3)/(x-2)
The domain is {x ∈ R and X ≠ 2}

Find the inverse function (1) y = 2x + 3 (2) y = log3 (x-3) (3) y = 3x + 1

1.x=2y+3
2.x=log3(y-3)
3.x=3y+1
To put it bluntly, it's the transposition of X and y

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F (x) = (a * 2 ^ x-1) / (2 ^ x + 1) satisfies f (- x) = - f (x); f (0) = (A-1) / 2 = 0
A=1
f(x)=(2^x-1)/(2^x+1);x∈R
f-1(x)=log2 (1+x)/(1-x);x∈(-1,1)
f-1(x)>log2(1+x) ;log2 (1+x)/(1-x)>log2(1+x)
log2(1-x)

It is known that the function f (x) = x2-2tx + 1, X ∈ [2,5] has inverse function, and the maximum value of function f (x) is 8. Find the value of real number t

Because the function has inverse function, it is one-to-one corresponding in the definition domain. The symmetric axis of F (x) = x2-2tx + 1 is x = t, so t ≤ 2 or t ≥ 5
If t ≤ 2, the function is monotonically increasing in the interval [2,5], so f (x) max = f (5) = 25-10t + 1 = 8, the solution t = 9
5
If t ≥ 5, the function is monotonically decreasing in the interval [2,5], so f (x) max = f (2) = 4-4t + 1 = 8, the solution t = - 3
4, which is inconsistent with t ≥ 5
To sum up, the value of the real number T satisfying the question is 9
Five

Given the function f (x) = (3x + 2) / (x + a), its inverse function is itself. Find the value of real number a

The inverse function of F (x) is
x=(3f(x)+2)/(f(x)+a)
The results are as follows:
xf(x)+ax=3f(x)+2
f(x)=(2-ax)/(x-3)
Because the inverse function is itself
(2-ax)/(x-3)=(3x+2)/(x+a)
A = - 3

If the function f (x) = x ^ 2 + ax + 1 has an inverse function on [0,2], find the value range of real number a

If the function f (x) = x ^ 2 + ax + 1 has an inverse function on [0,2], then the axis of symmetry is not in the interval (0,2)
That is - A / 2 = 2, the solution a > = 0 or a