Question

If h(x) is the inverse function of f(x), find h'(5)

Answer 1

\[f\circ h(x)=x\\
f'(h(x))\times h'(x)=1\] by the chain rule

Answer 2

that means
\[h'(x)=\frac{1}{f'(h(x))}\]

Answer 3

of course that doesn't tell you what \(h'(5)\) is unless you know \(f(x)\)

Answer 4

it just gives me a table of values for each function f(x) 5,8; f'(x) 4, 6; f"(x) -3,10

Answer 5

well maybe that is enough
the numbers you need are \(h(5)\) and \(f'\) of that number

Answer 6

can you post a screen shot or something? this is a bit hard to read

Answer 7

yes just a second

Answer 8

ok i see it from the table
\[f(3)=5\]right?

Answer 9

yes

Answer 10

that means \(h(5)=3\) so now we are at
\[\frac{1}{f'(h(5))}=\frac{1}{f'(3)}\]

Answer 11

if you can find \(f'(3)\) in the table you are done

Answer 12

no those are the only numbers given so i only see f'(4)

Answer 13

hmm let me look again

Answer 14

oh no, you are definitely told what \(f'(3)\) is
look at the bottom row of the table

Answer 15

5?

Answer 16

is it not clear how to read the table?

Answer 17

oh. I see now. Thank you so much for your help!

Answer 18

oh wow sorry you are right, i was reading it wrong !

Answer 19

the top row has a 3, and so yes, \(f'(3)=4\)

Answer 20

that makes your answer \(\frac{1}{4}\)

Answer 21

got it thank you so much! any chance you have time for one more? I can take a screen shot of it as well

Answer 22

yes, a screen shot will help

Answer 23

#15. the one that is circled

Answer 24

you need the derivative of arctangent first
do you know it?

Answer 25

if not you can look it up

Answer 26

\[\frac{ 1 }{ 1+x ^{2}? }\]

Answer 27

yes, and so
\[f'(\sqrt{3})=\frac{1}{1+\sqrt{3}^2}=\frac{1}{4}\]

Answer 28

that is the slope of your line
now you need the point on the graph of \(f(x)=\arctan(x)\) and \(x=\sqrt3\)
do you know what \(\arctan(\sqrt3)\) is ?

Answer 29

if not, i will tell you

Answer 30

.939?

Answer 31

oh my you need an exact answer, not a decimal
it is \(\frac{\pi}{3}\)

Answer 32

i just typed it in my calculator i didn't know it was supposed to be in radians

Answer 33

i just pulled up a unit circle and I see it now

Answer 34

yeah of course
in any case you can find the equation of the line with slope \(\frac{1}{4}\) through the point \((\sqrt3,\frac{\pi}{3})\)

Answer 35

use the point slope formula
\[y-\frac{\pi}{3}=\frac{1}{4}(x-\sqrt3)\]

Answer 36

right, so is k 3root3/12 plus 4pie/12?

Answer 37

i guess, i didn't do it but it is something ugly like that

Answer 38

ok i just wanted to make sure i wasn't supposed to compute that out into one single number with a decimal

Answer 39

no, i it is whatever you get when you write the line in standard form
multiply by 4 etc

Answer 40

ok great thank you. you were a lot of help

Answer 41

yw