Math Is Fun Forum

sarsalan

Member

Registered: 2008-11-03

Posts: 16

Real Valued Function Please help this is urgent

Question:   For the real valued function, f defined below     
                 Find  f^-1(-1)   
                 Where
                           f(x)=3x^2+7

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Real Valued Function Please help this is urgent

Clearly f(x) can never be less than 7 (because 7 is 7 and the square is positive).
The question is basically asking what value of x gives f(x) = -1, and for real values of x this never happens.

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Why did the vector cross the road?
It wanted to be normal.

Rozain

Guest

Re: Real Valued Function Please help this is urgent

Question:   For the real valued function, f defined below     
                 Find  f^-1(-1)   
                 Where
                           f(x)=3x^2+7

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Real Valued Function Please help this is urgent

Clearly f(x) can never be less than 7 (because 7 is 7 and the square is positive).
The question is basically asking what value of x gives f(x) = -1, and for real values of x this never happens.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Real Valued Function Please help this is urgent

Does f^-1 mean to take the reciprocal when x=-1 ?
If so, plug in x=-1, get 7+3 or 10, and flip it over as a fraction
and get 1/10.

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igloo myrtilles fourmis

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Real Valued Function Please help this is urgent

It's more likely to mean the inverse function.

ie. the function that makes f^-1(f(x)) ≡ x

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Why did the vector cross the road?
It wanted to be normal.

G_Einstein

Member

Registered: 2008-08-30

Posts: 124

Re: Real Valued Function Please help this is urgent

f^-1 is the invers function.

f(x)=3x^2+7   f(x)=y

y=3x^2+7
y-7=3x^2  /:3
(y-7)/3=x^2  / squareroot

sqrt[(y-7)/3]=x   so

sqrt[(x-7)/3]=y

sqrt[(x-7)/3]=f(x)

f(x)=sqrt[(x-7)/3]     is the invers function of f(x)=3x^2+7
Now f^-1(-1)

f(-1)=sqrt[(-1-7)/3]
f(-1)=sqrt[-8/3]
f^-1(-1) isn't defined because the result is a complex number,and function can not have complex solutions

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