Math help... Given f(x)=x^2 and g(x)=2^x?

Question

The inverse of g is a function, but the inverse of f is not a function. Explain why this statement is true.




Find g^-1 (f(3)) to the nearest tenth.

Answer 1

definition of a function: for every x there is one, and only one, y.
inverse of f is the sqrt which would have two values + and -.

To find g^-1:
y = 2^x becomes x = 2^y and solving for y:
log x = y log 2
y = log x / log 2

f(3) = 9

g^-1(9) = log 9 / log 2 = ~ 3.2

that's it! :)

Answer 2

LOOKING AT OTHER ANSWERS, I THINK I'M WRONG!!! (I'M ONLY A GCSE STUDENT) - sorry!!!

BUT THIS METHOD DOES SHOW YOU HOW TO FIND INVERSE FUNCTION REALLY EASILY

f(x)= x²
Lets say
y = x²
To calculate inverse function make all ys xs, and all xs ys
x = y²
Rearrange to y= (thats the inverse function)
y = square root (x)
So inverse f(x) is
sqroot x

g(x)=2
y = 2^x
x = 2^y
Now you can't rearrange to y= , so inverse of g(x) isnt possible


I think you got the question wrong; inverse f(x) is possible, but inverse g(x) ISN'T

Answer 3

f(x) does not pass the horizontal line test so its inverse is not a function. g(x) does pass the horizontal line test so its inverse is a function.

The inverse of g(x) is ln(x)/ln(2)
f(3) = 9
So g^-1(9) = ln(9)/ln(2) = 3.2

Answer 4

Ok, we begin by finding the inverses:

If f(x) = x^2

Then f^-1(x) = +/- sqrt(x) ; This equation fails the vertical line test, it is actually an ellipse.

If g(x) = 2^x

Then g^-1(x) = log_2(x) ; This equation passes the VLT and is a function.

f(3) = 3^2 = 9

g^-1(f(3)) = log_2(9)

g^-1(f(3)) = ln(9)/ln(2)

g^-1(f(3)) = 3.2

Answer 5

f(x) = x²
g(x)=2^x

g^-1(x)
Let g(x) = y
y=2^x
x=2^y (ln both sides)
lnx=yln2
y=lnx/ln2
g^-1(x) = lnx/ln2

Now inverse of f^-1(x) = x²
x=y²
y=√x
y=±√x
f^-1(x)=±√x
It is not a function , because for one value of y there are two values of x.

g^-1(x) = lnx/ln2
f(3)=(3)²=9
g^-1(f(3)) = lnx/ln2
g^-1(f(3)) = ln9 / ln2 = 3.169925001 ≈ 3.2