QUESTIONS - HOPE - FUNCTIONS
QUESTIONS - ANSWERS - FUNCTIONS
On 12/1/06, Hope
> Hope has left a new comment on your post "QUESTIONS- HOPE -
> INEQUALITIES,RATIO-PROPORTION":
>
> If two relations in x and y are inverses of one another, then across what
Ø line are they reflections of one another?
For the function
Y=f(x) , we consider x as independent variable and y as dependent variable.
The inverse function is derived from the same relation as above
Treating y as independent variable and x as dependent variable.
Hence they represent reflections along the line y=x
==============================================
Determine f^-1(x) for the
following function. f(x) = 10 (x – 1)
1.first let y = f(x)
let y=f(x) =10(x-1)=10x-10
2. next transpose and find x in terms of y.
10x = y+10
x = (y+10)/10
3.now the resultant equation is the inverse function…that is the
equation obtained for x on RHS is the f^(-1) function.
To put it in standard form replace y with x on RHS to get
F^(-1)[x]
hence
f^(-1) [x] = (x+10)/10
==========================================
Determine the solution to the
following composite function. (f^-1 ◦ f) (y)
to find
f^(-1)[f(y)]
let z = f(y)…….from f(x)=10x-10
z = 10y-10
f^(-1)[x]=(x+10)/10
f^(-1)[z] = (z+10)/10
but
z=10y-10…..hence
f^(-1)[f(y)]= (z+10)/10 = (10y-10+10)/10 = y
> Posted by Hope to Maths,management,counselling at 7:15 PM
2 Comments:
Hello. Do you mind telling me if I got these 2 correct? Thanks!
Solve the following for x = -3 and tell what kind of function it is.
0.1^x
-.001. This is an exponential function.
Solve the following for x = 3 and tell why ‘e’ is an irrational number.
e^x
20.0855 This is irrational because the number doesn’t terminate or repeat.
5:03 PM
Can someone please help me with this problem? (u,v) is the inner product general notation.
Let u=(u1,u2) and v=(v1,v2)
Show that
(u,v)=4u1v1+u2v1+u1v2+4u2v2
is an inner product on R^2 by verifying that the four inner product axioms hold.
Here are the four axioms:
axiom 1: (u,v)=(v,u)
axiom 2: (u+v,z)=(u,z)+(v,z)
axiom 3: (ku,v)=k(u,v)
axiom 4: (v,v)>= 0 and (v,v)=0 if and only if v=0.
4:59 PM
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