Help!!!!!! Can anyone explain how to solve these problems?

Question

Solve the equation for x using the given information:

a) 4 + ƒ(x+3) = -3; ƒ ֿ¹(-7) = 0

b) ƒ ֿ¹(x+1/x-2) = 12; ƒ(12) = 13

Answer 1

I won't solve it for you, but I can try help explaining what it means.

First of all, this is trying to help you understand something called inverse functions (http://www.uncwil.edu/courses/mat111hb/functions/inverse/inverse.html - try this site for more information)

Now, I'm not sure what level of maths you do know. But basically a function is a process that you undertake with a number. So, if I have a function which says any number I have (which I'll called x) I would multiply it by 3 i.e. f(x) = 3x. So, if we had the number: 5 (i.e. x =5) then f(5) = 3*5 = 15.

Now there is something called inverse function, this means I'm reversing the function i.e. since f(5) = 15 then the inverse has to be ƒֿ¹(15) = 5 (i.e. ƒֿ¹ is what we used to represent the inverse).

This means ƒֿ¹=x/3, therefore using x (which again means any number) in this case to be 15, you'll see we get back 5.

Now, the problems you have are a little more complicated but essentially the same problem.

f(x+3): just means for any number (again represented by x), we would first add 3 and then apply the function to it. In this case the function is not stated.

That is f(x+3) is the similar to saying f(2+3) = f(5). Now, I just chose x to be 2.

The problem wants you to be find out for what value of x, that 4 + f(x+3) = -3 is true given that the inverse function of -7, ƒֿ¹(-7) =0.

If you follow, what I've said before you should be able to do this easily.

The second one is similar, but if you might find it easier if you rewrite it as (since this looks just like the first question you had in a):
ƒ(x+1/x-2) = 12; ƒֿ¹(12) = 13

Answer 2

Before I start these, one thing to understand about one-to-one functions are that, if

f(x) = y, it follows that
f^(-1)(y) = x

Also, functions and their inverses cancel each other out.
Example:

ln(e^x) = x {ln and e^ are inverses of each other}
[x^(1/3)]^3 = x {cube root and cube are inverses of each other}

Therefore

f^(-1) (f(x)) = x, and
f(f^(-1)(x)) = x


a)

4 + f(x + 3) = -3, f^(-1) (-7) = 0

Rearranging the first expression,

4 + f(x + 3) = -3
f(x + 3) = -7

Since f^(-1)(-7) = 0, it follows that
-7 = f(0)

Therefore f(0) = f(x + 3), so
0 = x + 3, and
x = -3

b) f^(-1) [ (x + 1) / (x - 2) ] = 12, f(12) = 13

f(12) = 13 means
f^(-1)(13) = 12, so

f^(-1) [ (x + 1) / (x - 2) ] = f^(-1)(13)

Equate the insides of the function, and we get

(x + 1) / (x - 2) = 13. Now we solve as per normal algebra.

(x + 1) = 13(x - 1)
x + 1 = 13x - 13
14 = 12x, so
x = 14/12 = 7/6

Answer 3

Subtract 4 from each side of the first one to get two equations:

f(x+3) = -7
f^-1(-7) = 0

Do you understand what these mean?

Some function of x-3 is equal to -7.
The inverse function of -7 is equal to 0.

It should be obvious from that.

Once you've solved that, the second problem should be straightforward.

Answer 4

360/n - 16 = 120/n ==> 360/n - 16 + 16 = 120/n + 16 ==> 360/n = 120/n + 16 ==> 360/n - 120/n = 120/n + 16 - 120/n ==> 360/n - 120/n = 16 ==> (360 - 120)/n = 16 ==> 240/n = 16 ==> (n/16)*(240/n) = (n/16)*16 ==> 240/16 = n ==> n = 15

Answer 5

I can't imagine why you think I'd do your homework for you!

Answer 6

secret!!!!!!