Question

Find F inverse prime at a=4
F(x)=2x^2+3x^2+7x+4
I have the Derivative of the inverse already I just want to know how you know f'(4)=0..

Answer 1

The Derivative of the inverse is y'=1/(6y^2+6y+7)

Answer 2

First question, in the inverse replace all x's with 4 and solve...
what did you get for the inverse?

Answer 3

2x^3 + 3x^2 + 7x + 4 = 4
2x^3 + 3x^2 + 7x = 0
x(2x^2 + 3x + 7) = 0
The quadratic in the brackets has no solutions (discriminant is negative), so only solution is x = 0.

Answer 4

I did implicate differentiation to get the derivative of the inverse function so I just replaced all the x with y... But If I replace all the x's with 4 it becomes a huge number not 0 specifically 208.

Answer 5

Oh that makes sense okay so would I do it like that on all problems? Just make the problem = to whatever a =?

Answer 6

So (f^-1)(4) = 0

f ' (x) = 6x^2 + 6x + 7
f ' (0) = 7
1 / f ' (0) = 1/7

Answer 7

Thank you! What happens if x = 2 numbers what do I do then?

Answer 8

find the derivative and replace x with 2
then take the inverse

Answer 9

Thanks! Would you be able to help me with one more problem?

Answer 10

So let a = f(x). Then:
(f^-1) ' (a) = 1 / f ' [(f^-1)(a)]

Answer 11

sure

Answer 12

I am typing out the problem now it will be a few minutes sorry

Answer 13

ok

Answer 14

\[If h(x)=x+\sqrt{x} find h'(6)\]

Answer 15

Wow that turned out bad can you read it?

Answer 16

find the derivative of h(x)...plug in value of 6 for x
what did you get for the derivative

Answer 17

just use the rule for summations...

Answer 18

1+1/2x^-1/2

Answer 19

correct, now just switch x for 6

Answer 20

Since its raised to the -1/2 it turns out really weird..

Answer 21

that just means the square root is in the denominator of the fraction..