+118723 f(x)=e^(2-x)-e^2-x?
$$f(x)=e^{(2-x)}-e^2-x$$
let f(x)=y
$$y=e^{(2-x)}-e^2-x$$
Here is the graph
https://www.desmos.com/calculator/swkwcvrths
I don't know how to make x the subject but I can see that y maps to x one to one
and the function is continuous
so I can just swap x and y over
$$x=e^{(2-y)}-e^2-y$$
This is not very elegant I am afraid :(
+118723 f(x)=κ-e^(2-x)+x
Here is a graph of f(x)
https://www.desmos.com/calculator/rr5rg1wjd3
since the mapping of x to y and y to x are both 1 to 1 I can just let f(x)=y and then swap x and y over.
function
$$y=k-e^{(2-x)}+x$$
inverse function
$$\\x=k-e^{(2-y)}+y\\$$
I should make y the subject but I cannot see how to do that. 
+118723 ok I will take a look :))
For each of the functions below, state the domain and range, the restrictions, the intervals of increasing and decreasing, the roots, y-intercepts, and vertices.
1) f(x)=2x^2−8
2) f(x)=+√x - 2
3) f(x)=x+1/x-1
$$1) \;\;f(x)=2x^2-8$$
Any real number can be squared so x can be any real number
Domain: $$x\in R$$ (x is in the set of reals) $$(-\infty,\infty)$$
$$x^2$$ cannot be negative, it can be any positive number though. Same for $$2x^2$$
$$2x^2-8$$ cannot be any smaller than -8
Range: $$-8\le f(x) < \infty\qquad \qquad [-8,\infty)$$
Even before you went to all this bother you should have recognised that this is a concave up parabola.
It is a steeper version of y=x^2 And every point is dropped 8 units.
In other words, The axis of symmetry is x=0 (That is the y axis)
The vertex is (0,-8)
The y intercept is -8
It is decreasing when x
It is increasing when x>0 (the tangent has a positive gradient)
When f(x)=0
$$\\ 0=2x^2-8\\
0=x^2-4\\
0=(x-2)(x+2)\\
x-2=0\qquad or \qquad x+2=0\\
x=2\qquad \qquad or \qquad x=-2\\
$The roots are x=2 and x=-2$$$
I think that covers the first one.
Think about it and aske questions if you have any.
Have a go at the second one yourself and present what you think.
Here is the graph:
https://www.desmos.com/calculator/gmxuxxwtbv
I will talk about it with you then :)
