Question

check answer @hartnn

Answer 1

@Michele_Laino

Answer 2

since \(f\) is increasing, then \(f'\) is greater than zero, and \( (f^{-1})'\) is also greater than zero, so the inverse function is also increasing

Answer 3

Okay, would it be concave downwards though?

Answer 4

yes! I think so!

Answer 5

Could you give a possible example?

Answer 6

an example can be this:
\(y=f(x)=x^2\)

Answer 7

i think option C is correct one

Answer 8

Suppose we have function
\[\large \bf y=x^2=f(x)\]

Answer 9

I thought of that, but x^2 is not always increasing

Answer 10

inside the intrval \(x>0\), \(f\) is concave upword

Answer 11

interval*

Answer 12

but we can take x>0,then it is always increasing

Answer 13

k

Answer 14

so what is the inverse then?

Answer 15

sqrt x?

Answer 16

correct

Answer 17

draw its graph

Answer 18

that's concave down right?

Answer 19

yep

Answer 20

but increasing

Answer 21

correct

Answer 22

Okay, thank you both!

Answer 23

np :)

Answer 24

and sorry @Michele_Laino sir !

Answer 25

:)

Answer 26

no problem :) @mayankdevnani