Inverse of (1 + tanx)^3/2

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Gascoigne

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f(x) = (1 + tan(x))[sup:37pfknz6]3/2[/sup:37pfknz6] for -pi/4 < x < pi/2

Let f[sup:37pfknz6]-1[/sup:37pfknz6] denote the inverse function of f. Write an expression that gives f[sup:37pfknz6]-1[/sup:37pfknz6](x) for all x in the domain of f[sup:37pfknz6]-1[/sup:37pfknz6].

y = (1 + tan(x))^3/2

y^2/3 = 1 + tan(x)

y^2/3 -1 = tan(x)

??

I can't isolate x to get f[sup:37pfknz6]-1[/sup:37pfknz6](y), so I can't substitute x and y for f[sup:37pfknz6]-1[/sup:37pfknz6](x). Does this need calculus?



- Dylan
 


Have you heard of the arctangent (or inverse tangent) function?

 
Gascoigne said:
... I don't know how to use [the arctangent function] to find x
Click to expand...


Then you do not understand what the arctangent function does.

Look up its definition. Then apply it to both sides of your last equation.

 
Re:

mmm4444bot said:
Gascoigne said:
... I don't know how to use [the arctangent function] to find x
Click to expand...


Then you do not understand what the arctangent function does.

Look up its definition. Then apply it to both sides of your last equation.
Click to expand...
That's what I don't get - arctan(tan(x)) will get me x on the right by "undoing" tan(x), but I'll end up with something useless on the left.
 


If you have tan(x) = C, and you want to solve this equation for x, then the artangent function is used as follows.

x = arctan(C)

Does that make sense?

 
Re:

mmm4444bot said:


If you have tan(x) = C, and you want to solve this equation for x, then the artangent function is used as follows.

x = arctan(C)

Does that make sense?
Click to expand...
That's what my "that" was referring to: arctan(y^2/3-1) = x
then switch: arctan(x^2/3 - 1) = y

Is this right?
 
Gascoigne said:
... arctan(x^2/3 - 1) = y

Is this right?
Click to expand...


Our posts crossed in cyberspace.

Yes, this is correct, but it is clearer to put grouping symbols around exponents comprised of more than one character.

f^-1(x) = arctan[x^(2/3) - 1]

 
Re:

mmm4444bot said:
Gascoigne said:
... arctan(x^2/3 - 1) = y

Is this right?
Click to expand...


Our posts crossed in cyberspace.

Yes, this is correct, but it is clearer to put grouping symbols around exponents comprised of more than one character.

f^-1(x) = arctan[x^(2/3) - 1]
Click to expand...
I should also [sup:2zljpltv]raise it visually[/sup:2zljpltv] since the bb tag is available, but... one can only do so much. :)

Thanks.
 
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