Question

Show that f(x)= (see attached) is one-to-one and find (f^-1)'(0) (f^-1 signifies inverse).

Answer 1

\[ \int_{2}^{x} \sqrt{1+t^2} dt \] ?

Answer 2

Yes

Answer 3

Yes

Answer 4

well, showing that it's one to one and the derivative of its inverse at x=0

Answer 5

what shall we do for one-to-one?

\(f(x)=f(y)\) -----> \(x=y\) am i right?

Answer 6

Yes, that makes sense

Answer 7

well one-to-one just means that there cannot be multiple x-values at one y-value

Answer 8

I feel like I should take the derivative and solve for x but this in an integral which confuses me

Answer 9

first try to one to one

ok let suppose \(f(x)=f(y)\) or

\[\int_{2}^{x} \sqrt{1+t^2} dt=\int_{2}^{y} \sqrt{1+t^2} dt\]\[\int_{2}^{x} \sqrt{1+t^2} dt-\int_{2}^{y} \sqrt{1+t^2} dt=0\]\[\int_{x}^{y} \sqrt{1+t^2} dt=0\]

Answer 10

make sense?

Answer 11

Yes... what rule are you applying from step 2 to step 3 out of curiosity

Answer 12

Additivity of integration on intervals

Answer 13

Okay so continue

Answer 14

so \(x=y\)

Answer 15

and function is one to one

Answer 16

And how would I go about getting the inverse of an integral... is it just switch the interval and add a negative in front?

Answer 17

emm...i think we cant calculate \(f^{-1} (x)\) directly from an integral...

but we can find \(f^{-1} (0)\) and so \((f^{-1})'(0)\)

Answer 18

Okay can you show me how to do that

Answer 19

ok befor that u must know that :

\[\text{Derivative of the Inverse Function : } \ \ \ (f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}\]

Answer 20

and what is \(f'\) here?

Answer 21

isn't the derivative of an integral just the function inside the integral

Answer 22

so sqrt (1+t^2) I would think

Answer 23

yeah libniz rule

Answer 24

@rymdenbarn

waiting for when u r online

Answer 25

I got it thanks

Answer 26

very well....:)