The inverse function of the function y = (x + 1) 2 + 1 (x ≤ 1) is How is it calculated

The inverse function of the function y = (x + 1) 2 + 1 (x ≤ 1) is How is it calculated

Inverse function of y = (x + 1) ^ 2 + 1:
First transformation: x = ± √ (Y-1) - 1
Because x ≤ 1, the inverse function is y = ± √ (x-1) - 1 (x ≤ 5)

What is the parity of the inverse function y equals to the x power of e minus the negative x power of e divided by 2?

Hello!
y =f(x) = [e^x - e^(-x)] /2
Definition field x ∈ R
f(-x) = [e^(-x) - e^x] / 2 = -f(x)
/ / F (x) is an odd function
Therefore, the inverse function of F (x) is also an odd function

What is the inverse function of y = (x power of 3 minus - x power of 3) / 2?

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§(sec^2x-1)dx=§d(tanx)-§dx=(1/2)tan^2x-x+C

Indefinite integral of square of TaNx

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Images of reciprocal functions are symmetric with respect to the line y = X
Proof: let the function be y = f (x)
Then the inverse function is x = f (y)
Let (m, n) be a point on the image of function y = f (x), then n = f (m)
Then the symmetric point of this point with respect to y = x is (n, m)
In this paper, the symmetric point is introduced, M = f (n) conforms to the inverse function x = f (y)
So the point (n, m) is on the inverse function
So y = f (x), x = f (y) is symmetric with respect to the image y = X
Essence: choose any point to prove, because of the arbitrariness of the point, it also has generality

How to draw the inverse function of tax and TaNx

The image of tangent function is like an old cinema chair. Its period (minimum) is π. The image of arctangent function is a single curve that seems to grow slowly. It is not a periodic function

Simplifying cos (2tan-1x) tan-1x refers to the inverse function of TaNx

Let Tan α = x, then tan-1x = α
Moreover, Tan (2 α) = 2tan α / (1 - Tan 2 α)
sec²α = 1 / cos²α = 1 + tan²α
　　cos(2tan-1x)
　　= cos(2α)
　　= cos²α - sin²α
　　= cos²α (1 - tan²α)
　　= (1 - tan²α) / (1 + tan²α)
　　= (1 - x²) / (1 + x²)

Cos (2x + y) = 3cosy, find TaNx * Tan (x + y)

cos(2x+y)=3cosy
cos(x+y+x)=3cos(x+y-x)
cos(x+y)cosx-sin(x+y)sinx=3[cos(x+y)cosx+sin(x+y)sinx]
2cos(x+y)cosx=-4sin(x+y)sinx
[sin(x+y)sinx]/[cos(x+y)cosx]=-1/2
tanx*tan(x+y)=-1/2

It was found that 1-2cosxcosx / cos square x-sin square x = 1-tan / 1 + TaNx

Proof: (1-2sinxcosx) / (COS? X-sin? X) = (1-tanx) / (1 + TaNx) = (1-2sinxcosx) / (COS? X-sin? X) = (COS? X-sin? X) / (cosx SiNx) (cosx +...)