How to graph any inverse functions

[noregret]

I am struggling with drawing graphing inverse functions. Can someone help with this please. For most questions, I cannot gralh inverse functions original functions on the same set of axes.

[Sine]

I am struggling with drawing graphing inverse functions. Can someone help with this please. For most questions, I cannot gralh inverse functions original functions on the same set of axes.

Are you able to get the rule of the inverse function easily?

Graphing is easier if you use the features of the original function such as intercepts. If the original function has an x intercept at (1,0) for example the inverse will have a y intercept at (0,1). Likewise if there is an y-intercept at (0,3) for the original function the inverse will have ea x-intercept at (3,0). While these are the "easier" points to take note of this can be done with any point as (x,y) on the original graph becomes (y,x) on the inverse function. This can be also done with the gradient but less precisely, i.e if there is a gradient of zero (think of when we have half of a quadratic and it's turning point) it's inverse will be a square root function which begins with a gradient approaching infinity.

Also it pays to dot a line of y=x if you are given grid lines (you get these in VCAA exams) , this follows on with the first technique, if the original function passes through y=x the inverse will pass through the same point. This helps when solving for intersections of the original function and it's inverse as you merely need to solve f(x)=x or f^-1(x)=x not f(x)=f^-1(x) (which sometimes gets very difficult to solve).

Hope this helps 

[M909]

You have to reflect the original graph in the line y = x . That means many point that has the same x and y value/as on line y = x (e.g. (1,1), (pi,pi), (505,505)) will be the same/intercepts between graph of original and inverse. For all other points, the coodinates reverse, e.g. (1,2) becomes (2,1) (5,7) becomes (7,5). If the original function is not a 1 to 1 function (i.e. only 1 y value for each x value and vise versa), then you'll have to restrict the original function, as i've done in attached sample q.

Also, to find inverse on CAS,(in my example i used original as f1(x) = x^2 +5x-2) you can sketch the original on graph (so it will be f1(x) = ..., or whatever number you're up to) , then open calculation, and in solve function type x = f1(y), y. You'll get one answer if original was one to one, but if not (as in my example attached), you'll be two. Hope this helps, good luck

[noregret]

Thanks for your reply, I can find the inverse rule easily. Can you tell me how to find f^-1(x)=f(x)

[noregret]

Thanks for your reply.

[Sine]

The thing is you won't need to find f^-1(x)=f(x) (it is solvable but is just will usually take more time/effort to get to the desired result) unless a question specifically states to do so. Thats why you should just use f(x)=x or f^-1(x)=x to yield the x coordinate for the intersection between f^-1(x) and f(x). Since all the intersection will be on the line y=x the y-coordinate for the said intersection will be the same number as the x-coordinate.

[Wota]

are inverses of a hyperbola always the same as the original hyperbola? eg. f(x)=1/x and inverse of f(x)=1/x?

[Shadowxo]

are inverses of a hyperbola always the same as the original hyperbola? eg. f(x)=1/x and inverse of f(x)=1/x?

Only if it's symmetrical along the line y=x.
eg it is the same as the original hyperbola if f(x)=1/x, f(x) = 1/(x+2)-2
if f(x)=a/(x-h) +h, then f-1(x)=f(x)