Question

Derivative question

Answer 1

POWER RULE

Answer 2

its always power rule

Answer 3

f(x) and g(x) are inverses.
\[f(x) = 4x^3 - \frac{ 7 }{ x } + 5\]
Find g'(8)

Answer 4

called it

Answer 5

I know that g'(x) = 1/f'(g(x))
I'm not quite sure what to put in for the g(x) part.. how do I find g(8) ?

Answer 6

Power rule here? lego are you sure? :D

Answer 7

just plug in 8 for x in your derivative

Answer 8

oh sorry I didnt fully read the question. one sec

Answer 9

ahh i knew it !!

Answer 10

oh no i didn't
you can usually find it with your eyeballs in these, not so much ere

Answer 11

Lol xD I did some complicated thing and my teacher said I was over complicating it... that there was a simple thing to plug in for g(8)? The bell literally rang when she was mid sentence -_-

Answer 12

usually they are cooked up so you can guess
like x = 1 or x = 2

Answer 13

hmmm

Answer 14

No easy way to do this one then?

Answer 15

lol try -1

Answer 16

:/

Answer 17

it is really guess and check

Answer 18

you are not going to solve a cubic
\[4-7+5\neq 8\] but \[-4+7+5=8\]

Answer 19

@abbles remember that since f and g are inverses, g(8) really means... f(x) = 8, you'd need to find x... \[\large 8 = 4x^3 - \frac{ 7 }{ x } + 5\]

Answer 20

If that doesn't make sense, since they're inverses...

first the definition of inverses:
If \(\large y = f(x) \) then \(\large f^{-1} (y) = x\)

And we know g is the inverse of f,
so \(\large g(8) = f^{-1} (8) \) right @abbles?


So then that means, from looking at the above definition of inverses,
\(\large f^{-1} (8) = x\) or \(\large 8 = f(x)\)

If that all makes sense.

And @satellite73 already showed that x=-1.

Answer 21

\[\large\rm 8=4x^3-\frac7x+5\]Multiply through by x,\[\large\rm 8x=4x^4-7+5x\]Subtract 8x to the other side,\[\large\rm 0=4x^4-3x-7\]Maybe factor out a 4,\[\large\rm 0=x^4-\frac34x-\frac74\]

Apply rational root theorem:
If a rational root exists, it must be a factor of our constant term -7/4.
So there are only a handful of numbers you need to check:
-1,1
-7,7
-1/4, 1/4
-1/2, 1/2
-7/2, 7/2
I know, it's still a lot :P But turns out x=-1 works out as was already mentioned! :D

So you discovered that \(\large\rm f^{-1}(8)=g(8)=-1\).

Use your formula for inverse derivative that you mentioned before,\[\large\rm (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}\]Evaluated at 8,\[\large\rm (f^{-1})'(8)=\frac{1}{f'(\color{orangered}{f^{-1}(8)})}\]So,\[\large\rm (f^{-1})'(8)=\frac{1}{f'(\color{orangered}{-1})}\]

Oh I guess you were just having trouble with the first part :P finding the root value. I see.

Answer 22

Thanks everyone who answered! So are most questions like these guess and check?

Answer 23

You can always use a calculator to solve \(\large 8 = 4x^3 - \frac{ 7 }{ x } + 5\)
Subtract 8 from both sides, then graph the equation and solve where it's equal to zero.