Question

Find F inverse prime at a=4
f(x)=2x^3+3x^2+7x+4

Answer 1

\[(f^{-1})\prime(4)\]

Answer 2

\[f(x)=2x ^{3}+3x ^{2}+7x+4\]

Answer 3

let:
\(\large y=2x^3+3x^2+7x+4 \rightarrow x=2y^3+3y^2+7y+4 \)
can you do implicit differentiation to get y' ???

Answer 4

I use implicit diferentiation okay I will try

Answer 5

But I get confuesed on how to get y by itself

Answer 6

you mean y' by itself... ???

Answer 7

yes that is what I mean

Answer 8

ok... let's work this out...

Answer 9

I can do the implicit part

Answer 10

\(\large x=2y^3+3y^2+7y+4 \)
\(\large 1=6y^2y'+6yy'+7y' \)

factor out a y' on the right hand side:
\(\large 1=(6y^2+6y+7)y' \)

Answer 11

oooh

Answer 12

so \(\large y'=\frac{1}{6y^2+6y+7} \)

Answer 13

now this is the derivative of the INVERSE of f....

Answer 14

oh okay

Answer 15

now there is some formula right?

Answer 16

so can you use this to answer your question?

Answer 17

hello?

Answer 18

I am figuring it out

Answer 19

i'm sorry.. my internet is being stupid....

Answer 20

since you know \(\large f^{-1}(4)=0 \), just plug in y=0 into the derivative:
\(\large y'=\frac{1}{6y^2+6y+7} \)....

Answer 21

The slope of the line tangent to the inverse function at point a, is simply the reciprocal of the slope of the tangent to the function at a.