Question

If f(x) =4x-(1/x) and (x>0) and g is the inverse of f then what is the value of g'(0)?

Answer 1

\[f(x)=4x-\frac{ 1 }{ x }\]
the first thing i did was look for the inverse so I could take the derivative.

Answer 2

\[x= 4y -\frac{ 1 }{ y }\]\[xy=4y ^{2}-1\]\[xy+1=4y^2\]\[y=\sqrt{\frac{ xy+1 }{ 4 }}\]
That's what i got for the inverse, is this correct?

Answer 3

Yes

Answer 4

So now all i do is find the derivative and plug in 0 right?

Answer 5

Yes, could you show your work?

Answer 6

okay.

Answer 7

\[\frac{1 }{ 2 }\left( \frac{ xy+1 }{4 } \right)^\frac{ -1 }{ 2 } \]

Answer 8

that's the power rule so i start multiplying by the derivative of the inside using the quotient rule

Answer 9

I have to go, but yes I believe the process you're doing is right. Gl

Answer 10

I'll let v=xy+1/4
\[\frac{ (4*\frac{ dy }{ dx }v)-(v*0) }{ 4^2 }\]

Answer 11

thanks

Answer 12

Yw

Answer 13

I've ended up with \[\frac{ 1 }{ 2 }\left( \frac{ xy+1 }{ 4 } \right)^\left( \frac{ -1 }{ 2 }\right) *\frac{ y+x \frac{ d }{ dx } }{ 4 }\]

Answer 14

Using your initial idea, the inverse is
$$
xy=4y^2-1\\
4y^2-xy-1\\
g(x)=y=\cfrac{x\pm \sqrt{x^2+16}}{8}
$$
Since \(x>0\), the inverse \(g(x)>0\).

The derivative of this inverse is
$$
g'(x)=\cfrac{1}{8}\left ( \cfrac{x}{\sqrt{x^2+16}}+1\right )
$$
Then \(g'(0)=\cfrac{1}{8}\).