2. Given f(x) = k(2+x), find the value of k if f^-1 (-2) = -3

3. Consider the following function:
f(x) = -2 all square (-x + 4) -3 find the equation of the inverse function

One you graph f(x) and its inverse on a grid

find Df, D f-1, Rf, R f-1

I will try to decipher.


2. f(x)=k(2+x), \;\ \text{find k if} f^{-1}(-2)=-3


Replace f(x) with y:

y=k(2+x)

swap x and y:

x=k(2+y).

Solve for y.

Then, sub in x=-2, set the equation equal to -3 and solve for k.



3. Find the inverse of f(x)=-2(4-x)^{2}-3

Replace f(x) with y:

y=-2(4-x)^{2}-3

swap x and y:

x=-2(4-y)^{2}-3

Now, solve for y.

They are easy to graph with a calculator or graphing utility.

Here is a link to a wonderful FREE graphing utility:

www.padowan.dk

I do not know what Df, Rf, and so forth means.

Galactus, I need to tell you something

2. Consider the following function f(x) = -2 sqrt (-x + 4) - 3
a) find the equation of the inverse function and graph f(x) and its inverse on a grid
b) find Df
Find D f^-1
Rf
R f^-1
Those are domain and range can you please show me the full work I've been having a lot of trouble on this thanks

Can you draw the graph of f(x) ?

The domain is the horizontal interval of a function.

Let's say that I had f(x)\ =\ x^2 for x\ \geq\ 0

Then, f^{-1}(x)\ =\ \sqrt{x} for x\ \geq\ 0

When you draw f and f^-1, you'll see that:

The domain of f will lie between 0 and infinity. That is looking on the horizontal axis, f starts at zero and never ends.

Similarly, the domain of f^-1 will be between 0 and infinity.

The range of f is the vertical interval. Looking on the y axis, f starts at 0 and rises to infinity. The range is thus 0 to infinity. Same thing for f^-1.

Now, can you deduce the range and domain of the functions in your problem?