do I have to be able to do this by hand?

[wombifat]

for questions asking for the point of intersections between the function and its inverse

eg. 2loge(x=3)+4 find the inverse function and the coordinates of the points of intersection of the graphs f(x)=y and y=f^-1(x)

can we just do that using the calculator? Or do we need to know how to do this by hand?

[Flaming_Arrow]

you can do it on the calc

[wombifat]

so it won't come up on the tech free exam?

[TrueTears]

If the inverse intersects with original. You can always do the simultaneous equation with y = x. (As the inverse and original will intersection on the line y = x)

There will be a high probability a question like this might appear in the tech free.

[Mao]

If the inverse intersects with original. You can always do the simultaneous equation with y = x. (As the inverse and original will intersection on the line y = x)

There will be a high probability a question like this might appear in the tech free.

that is partially correct. For this method to work, the graph must not be symmetrical about

consider the graph , or a quarter circle . There are infinite number of solutions to

Hence, graphing is very important. Though for the sake of tech-free exam one, the method TT described would be sufficient [as they are not going to give you a psycho function that requires a calculator to solve for intersections not on y=x]

[NE2000]

If the inverse intersects with original. You can always do the simultaneous equation with y = x. (As the inverse and original will intersection on the line y = x)

There will be a high probability a question like this might appear in the tech free.

Yeah but is there any way you could do that on tech free with wombifat's y = 2loge(x-3) + 4??

[wombifat]

so would you find the original graph's intersection with y=x?

[TrueTears]

You can use either the original or the inverse.

And, just like Mao said, do a quick sketch of the graph to get an idea of what it looks like.

[wombifat]

yeah sketching a graph was part of the question anyway
thanks a lot, I never got told how to do questions like that without a calculator