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Given that the inverse function of the function y = f (x) is y = F-1 (x), then the expression of the inverse function of the function y = 2F-1 (3x + 4)
y=2f^(-1)(3x+4) y/2=f^(-1)(3x+4) f(y/2)=3x+4 x=[f(y/2)-4]/3
x. Y interchanges y = [f (x / 2) - 4] / 3
Given that the inverse function of the function y = 3x + 1 (x + a) {(x + a) (3x + 1)} is itself, then a=
The inverse function of y = 3x + 1 (x + a) is itself
Then we first find its inverse function
y=(1-ax)/(x-3)
Because the two equations are the same,
So compare the items
-a=3
a=-3
[or another way: because the inverse function is itself, so the point (x, y) and the point (y, x) are on this function, then there is
(x+a)y=3x+1---1
(y+a)x=3y+1---2
So subtracting the two equations 1 and 2 gives you
a(y-x)=3(x-y)=-3(y-x)
We get a = - 3]
Given f (x) = 3x + 5, find the inverse function ↓ It is known that f (x) = 3x + 5 (1) Find f ^ - 1 (x) (2) Find f ^ - 1 [f (x)] (3) Find f [f ^ - 1 (x)] Please write down the specific steps! Thank you! 00
(1)y=3x+5
x=(y-5)/3
f^-1(x) =(x-5)/3
(2)
f^-1(f(x))=f^-1(3x-5)=[(3x-5)-5]/3=x
(3)
f[f^-1(x)]=f[( x-5)/3]=3*[(x-5)/3]-5=x
How to do the inverse function of F (x) = 3x + 1?
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According to the meaning of the title, f ^ - 1 (x) = log2 is the bottom x (x > 0)
1) . f ^ - 1 (x) 0 so
X>0
2) When. X > 0, the range of 3x? 2 + x > 0 h (x) = log2 is the base (3x? 2 + x) and the value range is - infinite to + infinite, and the inverse function is h (x) = big root (2 ^ y + 1 / 12) / root sign 3 - 1 / 6
I don't know if it's wrong. Ha ha
Find the inverse function of the function y = log4 base 2 + log4 base radical x fast
Definition domain x > 0
y=log4(2)+(1/2)+[log4(x)]/2
y=1/2+(1/2)[log4(x)]
log4(x)=2y-1
x=4^(2y-1)
Inverse function y = 4 ^ (2x-1) x ∈ R
Inverse function of function f (x) = 3x + 2 / x + A
The inverse function is y = 2-ay / Y-3
Because the original function is y = 3x + 2 / x + a
So XY + ay = 3x + 2
xy-3x=2-ay
So y = 2-ay / Y-3
How to calculate the inverse function of F (x) = 1-3x / X-2
It should be f (x) = (1-3x) / (X-2)
f(x)*(x-2)=(1-3x)
The solution is: x = [1 + 2F (x)] / [f (x) + 3]
The inverse function
f(x)'=(1+2x)/(x+3)
How to calculate the inverse function of y = 3x-6
3x=y+6
x=y/3+2
The inverse function is
y=x/3 +2
Find the inverse function of function y = (1-3x) / (X-2)
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