Question

If f(x)=x^3+ e^(x/2) and g(x)=f inverse x
then find g '(1)

Answer 1

@experimentX

Answer 2

use the same technique

Answer 3

I did..ill tell you where I am stuck exactly

Answer 4

put x=1, that's all

Answer 5

swithc y and x and solve for y and u get inverse function

Answer 6

\[ \LARGE g'(x)=\frac{1}{f ' g(x))}\]
now putting 1
\[ \LARGE g'(1)=\frac{1}{f ' g(1))}\]
how do i get the value for f'(g(1))?

Answer 7

differentiate f(x) and put x = g(x)

Answer 8

\[\LARGE f'(x)=3x^2+\frac{1}{2}e^{\frac{x}{2}}\]

Answer 9

\[\LARGE f'(g(1))=3(g(1))^2+\frac{1}{2}e^{\frac{g(1)}{2}}\]

Answer 10

nowhere near the answer :O

Answer 11

first off, \(g^{-1}(1)=0\)

Answer 12

so your job it to compute
\[\frac{1}{f'(0)}\]

Answer 13

\[\LARGE g'(1)=\frac{1}{3(g(1))^2+\frac{1}{2}e^{\frac{g(1)}{2}}}\]
isnt the answer this?

Answer 14

i.e. take the derivative of \(f\)
then compute \(f'(0)\)
then take the reciprocal

Answer 15

hell no

Answer 16

forget the inverse function, you do not need it

Answer 17

@experimentX :O

Answer 18

these are the steps
1) first notice that \(f^{-1}(1)=0\)
2) compute \(f'\)
3) evaluate \(f'(0)\)
4) take the reciprocal of it

Answer 19

\[\LARGE g'(1)=\frac{1}{f '( g(1))}=\frac{1}{f '(0)}\]

Answer 20

okay f(0)=1 so f inverse 1 =0
f'(x) is already calculated
f'(0)=1 i guess :O
1/1=1? :|

Answer 21

f'(0)=1/2 sorry

Answer 22

i think your derivative may be wrong
ok now it is right

Answer 23

so your answer is \(2\)

Answer 24

$$f(x)=x^3+ e^{x/2}\implies f'(x)=3x^2+\frac12 e^{x/2}\\g(1)=x\implies f(x)=1\\x^3+e^{x/2}=1\implies x=0\\g'(1)=\frac1{f'(0)}=\frac1{3(0)^2+\frac12 e^{0/2}}=\frac1{\frac12}=2$$

Answer 25

(just supplementary info to go along with satellite73's explanation)

Answer 26

thanks everyone :) can anyone tell me what my solution is trying to tell me though?

Answer 27

g(1) = 0 everything is all right

Answer 28

\[f(\alpha)=1\]
\[\alpha^3+e^{\frac{\alpha}{2}}=1\]
comparing..
alpha=0
f(0)=1
f inverse 1 =0
i know the final outcome is correct but why did we take alpha n stuff?

Answer 29

Your solution is the slope of the inverse of \(f\) where \(f=1\).

Answer 30

@DLS we need to work backwards to figure out where (i.e for what \(x\) or \(\alpha\)) we get \(f=1\). As it turns out, you can solve this one by "inspection" pretty easily.

Answer 31

so everytime i substitute x=0 and see the value of f(0) ?
what if it gives no clue?

Answer 32

for one thing, it tells you if you know the slope of \(f\) at a point \((a,b)\), you also know the slope of \(f^{-1}\) at \((b,a)\)

Answer 33

how did we get the value for g(1)?

Answer 34

\[\Large g(1)=f^{-1}(1)\]
so am i supposed to see at what value f gives 1?

Answer 35

yuhuuuu @oldrin.bataku

Answer 36

just plot if you find it hard

Answer 37

so we are supposed to do that only?

Answer 38

yes you needed to find g(1) in your above equation.
if inverse can be called function at x=1, then defintely you can find the value of invsere. luckly it turned out to be zero.

Answer 39

great! question wont demand anything above 0,1,2 etc until its a very rare case..i hope?

Answer 40

the question are carefully constructed ... i hope they won't