Question

Find $g(1)$ where $g(x)$ is the inverse of $f(x)=\sin(x)-2x+1$

score 53 · Answer 1 · 2026-03-26 11:01:37

53

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Find $g(1)$ where $g(x)$ is the inverse of $f(x)=\sin(x)-2x+1$

Published on 26 Mar 2026 - 11:01

score 57 · Answer 2 · 2026-03-27 14:55:10

57

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How to derive this matrix formula with "vector inversion" which sometimes works and sometimes doesn't?

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score 139 · Answer 3 · 2026-03-28 06:07:52

139

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Positive real solution to $x^n=x+1$

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score 64 · Answer 4 · 2026-03-26 07:38:33

64

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When dealing with matrix exponentials, can we assume $e^{-A}=(e^A)^{-1}$?

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score 79 · Answer 5 · 2019-07-27 17:47:16

79

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Finding the right-inverse of a function

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score 74 · Answer 6 · 2026-03-26 17:30:16

74

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score 123 · Answer 7 · 2019-07-30 00:50:19

123

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If the left-inverse matrix is multiplied to the right of the matrix, why would the result be an zeros added identity matrix?

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score 79 · Answer 8 · 2026-03-27 18:27:53

79

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How to find inverse of $y = x C_1 + \tan^{-1}(x + rh) + C_2$?

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46

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what property between two matrices affects on the trace of multiply of them

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score 151 · Answer 15 · 2026-03-26 17:11:28

151

Views

Why does $A^{-1} = \frac{1}{|A|} \cdot \operatorname{adj}(A)$ work, intuitively?

Published on 26 Mar 2026 - 17:11

List Question

Find $g(1)$ where $g(x)$ is the inverse of $f(x)=\sin(x)-2x+1$

How to derive this matrix formula with "vector inversion" which sometimes works and sometimes doesn't?

Positive real solution to $x^n=x+1$

When dealing with matrix exponentials, can we assume $e^{-A}=(e^A)^{-1}$?

Finding the right-inverse of a function

Multiplicative inverses why two z in this formula?

If the left-inverse matrix is multiplied to the right of the matrix, why would the result be an zeros added identity matrix?

How to find inverse of $y = x C_1 + \tan^{-1}(x + rh) + C_2$?

Looking for a function having its fourier components

How to prove the surjectivity of an operator

Invertibility (and eigen-decomposition) of complex symmetric matrix?

Affine cipher easy question

what property between two matrices affects on the trace of multiply of them

Inverse function as series

Why does $A^{-1} = \frac{1}{|A|} \cdot \operatorname{adj}(A)$ work, intuitively?

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