Inverse Functions Questions (1 Viewer)

[.ben]

Find functions which are the inverse of each of the following. State the domain and range of the inverse in each case.

a) f(x)=2^-x, x>0

b) f(x)=x²+2x, x_>0

for the first one i got f^-1(x)=(-lnx)/(ln2) with D: x>0, R: y>0

and the second one i got f^-1(x)=-1+sqrt(x+1) with D: x_>-1, R: y_>0

thanks

 

[word.]

1.
y = 2^(-x), x > 0
at x = 0, y = 1
as x -> infinity, y -> 0
so the range is 0 < y < 1

f^-1(x) = -ln(x)/ln(2)
from the original function,
domain: 0 < x < 1
range: y > 0


2.
y = x^2 + 2x, x > 0

x = y^2 + 2y
x = (y + 1)^2 - 1
y = +-Sqrt(x + 1) - 1
f^-1 = Sqrt(x + 1) - 1 (original function domain x > 0 implies f^-1 > 0)

from the original function,

range: y > 0
i.e. f^-1 > 0
so Sqrt(x + 1) - 1 > 0
Sqrt(x + 1) > 1
so domain: x > 0

 

[.ben]

oh bugger, i get it now. i think it helps to draw the diagram now actually thanks.

 

[darkliight]

Very minor, but just in case your teacher knocks you back for it, a) has a range of 0 < y <= 1 and the inverse a domain of 0 < x <= 1.

 

[.ben]

yep tanks

 

[word.]

the range is the set of all possible y values for given domain
y = 2^(-x) = 1/2^x

the domain of the function is x > 0
since the function is monotonic (decreasing), we can expect the extremes of the range to correspond to the extremes of the domain (0 to infinity)

by checking the limiting behaviour as x approaches 0 and infinity we get our range
as x -> 0, y -> 1; x -> infinity, y -> 0,

so the function is encapsulated between the y values of 0 and 1
so the range is 0 < y < 1

Very minor, but just in case your teacher knocks you back for it, a) has a range of 0 < y <= 1 and the inverse a domain of 0 < x <= 1.

since the domain is x > 0, x is never 0 i.e. y = 1/2^0 = 1,
the range is the open interval 0 < y < 1

 

[.ben]

o i have another 2 questions:

Find functions which are the inverse of each of the following. State the domain and range of the inverse in each case.

1) f(x)=sqrt(9-x²) for -3<=x<=0


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2) is it possible to find the domain and range of a function which you can't draw?

 

[darkliight]

since the domain is x > 0, x is never 0 i.e. y = 1/2^0 = 1,
the range is the open interval 0 < y < 1

Ofcourse, sorry about that.

 

[word.]

1.
f(x) = Sqrt(9 - x^2), -3 <= x <= 0
x = Sqrt(9 - y^2)
x^2 = 9 - y^2
y = +-Sqrt(9 - x^2)
f^-1 = -Sqrt(9 - x^2) (Dom(f): -3 <= x <= 0 -> Range(f^-1): -3 <= y <= 0)
so range: -3 <= y <= 0
at y = 0, x = +-3
at y = -3, x = 0
domain: -3 <= x <= 3

2.
domain and range of a function which you can't draw?
what kind of function can't you draw?
um,

we'll define a piecewise function with the conditions
f(x) = 1 for all rational numbers x
f(x) = 0 for all irrational numbers x
this function is undrawable

range: 0 <= x <= 1
domain: all real x

answer: yes it is

 

[word.]

also,
f(x) = sin(1/x)
domain: all real x, x != 0
range: -1 <= y <= 1

 

[.ben]

o cool thanks word.!

lol one last question:

Without the aid of tables find:

a) sin^-1sqrt(3)/2

are you allowed to use a calculator?

 

[pLuvia]

If it says "Without the aid of tables" that usually also means you can't use a calculator
y=sin^-1sqrt{3}/2
siny=sqrt{3}/2
y=pi/3

 

[Riviet]

are you allowed to use a calculator?

There's nothing stopping you from using it, but make sure you show the relevant working to show you didn't actually find the answer from doing it on the calculator. Of course you can use your calculator to check your answer quickly.

 

[.ben]

ok then lol. but the working seems so pintless!

 

[Riviet]

It was a fairly straight forward question anyway, provided you knew your exact values, meaning a calculator wouldn't be necessary.

 

[.ben]

lol i don't even know my exact values, i just use a calculator.