Question

Let f (x) = 7x + sqrt(x + 50)
.

What is the form of the largest interval on which f has an inverse
and the value(s) of the endpoint(s)
find (f ^−1)' (0)

Answer 1

@zepdrix @wio can you guys help?

Answer 2

Hmmmm.

Answer 3

I know how to find \(f^{-1}(0)\)

Answer 4

Basically it's just: \[
0= 7x + \sqrt{x + 50}
\]

Answer 5

Then multiply by \(7x-\sqrt{x+50}\).

Answer 6

So \[
0=(7x)^2-(\sqrt{x+50})^2
\]

Answer 7

so is f^-1(0)=(7x)^2-x+50?

Answer 8

That \(-\) should distribute to both terms.

Answer 9

As far as an inverse goes... you can't invert it for \(x\) values where there is another \(x\) value has the same output.

Answer 10

It's clear the lowest value of \(x\) where \(f(x)\) is defined is where: \[
x+50\geq 0
\]

Answer 11

So \[
x\geq -50
\]

Answer 12

your equation thing is not working. I dont understand what your writing.

Answer 13

Oh wait... is it asking for derivative of inverse function?

Answer 14

That makes so much more sense.

Answer 15

yeah I think so

Answer 16

also the interval

Answer 17

If you find when the derivative is 0, then you find the max.

Answer 18

max or min

Answer 19

so the derivative is 7+(1/2)(x+50)^(-1/2)

Answer 20

what is the interval?

Answer 21

Find out when the derivative equalis 0 first.

Answer 22

I got 1/196-50

Answer 23

What does \(f(x)\) equal at that point?

Answer 24

-349.89

Answer 25

I think its supposed to be -(1/196)-50

Answer 26

I think the interval would just be\[
\left[\frac{1}{196}-50,\infty\right)
\]

Answer 27

its not 1/196-50 but the interval is correct otherwise

Answer 28

its [-50, infinity)