Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$
$begingroup$
Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$
My try:
We have $$y=sec xleft(sec x+tan xright)$$
$$2y=left(sec x+tan x+sec x-tan xright)left(sec x+tan xright)$$
$$2y=left(sec x+tan xright)^2+1$$
So we get:
$$sec x+tan x=sqrt{2y-1}$$ Also
$$sec x-tan x-frac{1}{sqrt{2y-1}}$$
Adding above two results we get:
$$x=sec^{-1}left(frac{sqrt{2y-1}+frac{1}{sqrt{2y-1}}}{2}right)$$
Hence $$f^{-1}(x)=sec^{-1}left(frac{sqrt{2x-1}+frac{1}{sqrt{2x-1}}}{2}right)$$
How to find Range?
algebra-precalculus trigonometry optimization maxima-minima inverse-function
asked Jan 8 at 4:51
Umesh shankarUmesh shankar
2,61931219
$endgroup$
add a comment |
$begingroup$
Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$
My try:
We have $$y=sec xleft(sec x+tan xright)$$
$$2y=left(sec x+tan x+sec x-tan xright)left(sec x+tan xright)$$
$$2y=left(sec x+tan xright)^2+1$$
So we get:
$$sec x+tan x=sqrt{2y-1}$$ Also
$$sec x-tan x-frac{1}{sqrt{2y-1}}$$
Adding above two results we get:
$$x=sec^{-1}left(frac{sqrt{2y-1}+frac{1}{sqrt{2y-1}}}{2}right)$$
Hence $$f^{-1}(x)=sec^{-1}left(frac{sqrt{2x-1}+frac{1}{sqrt{2x-1}}}{2}right)$$
How to find Range?
algebra-precalculus trigonometry optimization maxima-minima inverse-function
asked Jan 8 at 4:51
Umesh shankarUmesh shankar
2,61931219
$endgroup$
add a comment |
$begingroup$
Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$
My try:
We have $$y=sec xleft(sec x+tan xright)$$
$$2y=left(sec x+tan x+sec x-tan xright)left(sec x+tan xright)$$
$$2y=left(sec x+tan xright)^2+1$$
So we get:
$$sec x+tan x=sqrt{2y-1}$$ Also
$$sec x-tan x-frac{1}{sqrt{2y-1}}$$
Adding above two results we get:
$$x=sec^{-1}left(frac{sqrt{2y-1}+frac{1}{sqrt{2y-1}}}{2}right)$$
Hence $$f^{-1}(x)=sec^{-1}left(frac{sqrt{2x-1}+frac{1}{sqrt{2x-1}}}{2}right)$$
How to find Range?
algebra-precalculus trigonometry optimization maxima-minima inverse-function
asked Jan 8 at 4:51
Umesh shankarUmesh shankar
2,61931219
$endgroup$
Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$
My try:
We have $$y=sec xleft(sec x+tan xright)$$
$$2y=left(sec x+tan x+sec x-tan xright)left(sec x+tan xright)$$
$$2y=left(sec x+tan xright)^2+1$$
So we get:
$$sec x+tan x=sqrt{2y-1}$$ Also
$$sec x-tan x-frac{1}{sqrt{2y-1}}$$
Adding above two results we get:
$$x=sec^{-1}left(frac{sqrt{2y-1}+frac{1}{sqrt{2y-1}}}{2}right)$$
Hence $$f^{-1}(x)=sec^{-1}left(frac{sqrt{2x-1}+frac{1}{sqrt{2x-1}}}{2}right)$$
How to find Range?
algebra-precalculus trigonometry optimization maxima-minima inverse-function
algebra-precalculus trigonometry optimization maxima-minima inverse-function
asked Jan 8 at 4:51
Umesh shankarUmesh shankar
2,61931219
asked Jan 8 at 4:51
Umesh shankarUmesh shankar
2,61931219
asked Jan 8 at 4:51
Umesh shankarUmesh shankar
2,61931219
asked Jan 8 at 4:51
Umesh shankarUmesh shankar
2,61931219
asked Jan 8 at 4:51
Umesh shankarUmesh shankar
2,61931219
2,61931219
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint:
For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$
Here $0<1-sin x<2iff f(x)>dfrac12$
If $y=f(x)=cdots,$
$$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$
answered Jan 8 at 5:00
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$
Here $0<1-sin x<2iff f(x)>dfrac12$
If $y=f(x)=cdots,$
$$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$
answered Jan 8 at 5:00
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Hint:
For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$
Here $0<1-sin x<2iff f(x)>dfrac12$
If $y=f(x)=cdots,$
$$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$
answered Jan 8 at 5:00
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Hint:
For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$
Here $0<1-sin x<2iff f(x)>dfrac12$
If $y=f(x)=cdots,$
$$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$
answered Jan 8 at 5:00
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
Hint:
For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$
Here $0<1-sin x<2iff f(x)>dfrac12$
If $y=f(x)=cdots,$
$$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$
answered Jan 8 at 5:00
lab bhattacharjeelab bhattacharjee
224k15156274
edited Jan 8 at 5:05
answered Jan 8 at 5:00
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 8 at 5:00
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 8 at 5:00
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
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