Inverse Functions Question (1 Viewer)

Shuuya · Apr 11, 2016

Could I get some help with (iii) please?

(This is Q7(a) from the 2002 3U HSC paper)

Thanks!

Paradoxica · Apr 11, 2016

Shuuya said:
Could I get some help with (iii) please?

(This is Q7(a) from the 2002 3U HSC paper)

Thanks!

You've already sketched g(x), which contains f(x) in it. Identify the component of g(x) which is f(x), and invert it.

InteGrand · Apr 11, 2016

Shuuya said:
Could I get some help with (iii) please?

(This is Q7(a) from the 2002 3U HSC paper)

Thanks!

$\noindent For the inverse function $y=f^{-1}(x)$, we have $y\leq 0$, and $x=e^y + \frac{1}{e^y}$. Now solve this for $y$ to get the solution, and take the solution with $y\leq 0$. To solve this, note that it becomes a quadratic in $e^y$ when we multiply through by $e^y$ and rearrange etc.$

Shuuya · Apr 11, 2016

InteGrand said:
$\noindent For the inverse function $y=f^{-1}(x)$, we have $y\leq 0$, and $x=e^y + \frac{1}{e^y}$. Now solve this for $y$ to get the solution, and take the solution with $y\leq 0$. To solve this, note that it becomes a quadratic in $e^y$ when we multiply through by $e^y$ and rearrange etc.$

That's what I got, but I'm not sure which solution is correct...

Paradoxica · Apr 11, 2016

Shuuya said:
That's what I got, but I'm not sure which solution is correct...

Test a single point to determine which branch is the inverse.

i.e. if (a,b) is a point on the (restricted) function, plug in x = b and take the sign which yields f^-1(x) = a

InteGrand · Apr 11, 2016

Shuuya said:
That's what I got, but I'm not sure which solution is correct...

$\noindent Recall $y$ needs to be less than or equal to $0$. Assuming you did everything correctly so far (didn't check over it), we need to take the solution that'll be less than 1, since to make a natural log be negative, the argument has to be less than 1. So, which of the solutions be the right one?$

Shuuya · Apr 11, 2016

Paradoxica said:
Test a single point to determine which branch is the inverse.

i.e. if (a,b) is a point on the (restricted) function, plug in x = b and take the sign which yields f^-1(x) = a

InteGrand said:
$\noindent Since $y$ needs to be less than or equal to $0$. So assuming you did everything correctly so far (didn't check over it), we need to take the solution that'll be less than 1, since to make a natural log be negative, the argument has to be less than 1. So, which of the solutions be the right one?$

The negative root is the solution right?

InteGrand · Apr 11, 2016

Shuuya said:
The negative root is the solution right?

$\noindent Correct! This is because $\frac{1}{2}\left(x-\sqrt{x^2-4}\right)$ will always be less than or equal to 1 in the domain of $f^{-1}$. (The other one is always greater than or equal to 1 in this domain.)$

Shuuya · Apr 11, 2016

Thanks for all the help

Inverse Functions Question (1 Viewer)

Shuuya

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