Question

g(x)= x^3 + 2x −1. What is the slope of the tangent line to the curve g^−1(x) at the point (2, 1)?

Answer 1

@Compassionate @tkhunny help plz?

Answer 2

If \(x = f(y)\), we have \(1 = f'(y)\dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{1}{f'(y)}\)

Since \(f^{-1}(x) = y,\;we\;have\;\dfrac{dy}{dx} = \dfrac{1}{f'(f^{-1}(x))}\)

So, \(f'(x) = 3x^{2} + 2\).

We also have \(f(2) = 1\) and (with a little effort) \(f(1) = 2\).

Thus: \(f^{-1}(2) = 1\)

For \(x = 2\), where does that leave us?

Answer 3

@tkhunny Trying to let it process, so the slope is 1?

Answer 4

You'll have to do better than that. Show ALL your intermediate results.

Answer 5

Sorry, im just trying to make it clear. We take the derivative and plug 2 for x?

Answer 6

#1 If you want to sound legitimate, never use the term "plug" again when you mean substitute or substitution.

We need the point (2,1), so yes, x = 2.

Show your intermediate steps.

Answer 7

f'(2) = 3(2)^2 + 2
= 14

Answer 8

Why do you care what f'(2) is?

\(f'(f^{-1}(2)) = f'(1)\)

Answer 9

We're supposed to be looking for f'(1) then? Blah, sorry. I am terrible at math.

f'(1) = 3(1)^2 +2
= 5

Answer 10

It makes my eyes go cross-eyed when dealing with inverses.

But we can try to keep things straight by remembering a few things:
(1) if you have a curve y= g(x) with points (x,y)= (x, g(x) )
then
(2) the inverse curve \(g^{-1}(x)\) is g(x) reflected about the y=x line
see the attached graph

(1) and (2) mean that if \( (x_0, y_0) \) is on the curve (x, g(x) ) then
\( ( y_0,x_0) \) is on the curve \(g^{-1}(x)\) and vice versa (i.e. swapping coordinates has the effect of reflecting that point about the y=x line)

Answer 11

as tkhunny showed, there is the general property
that for a point \( ( x_0, y_0) \) on the g(x) curve, we can compute the derivative (i.e. slope of the tangent line) at the point \( (y_0, x_0) {\bf \text{ on the }g^{-1}(x)\ curve} \)

in other words,
\[ m|_{y_0}= \frac{d g^{-1}}{dx} \left(y_0\right) = \frac{1} {g'(x_0)}\]

they give us (2,1) on the \(g^{-1}(x)\) curve, which means the "swapped coordinates" (1,2) is on the g(x) curve, and
\( x_0= 1 , y_0= 2\)

we use that to find the slope at the point (2,1) on the inverse function curve is
\[ m= \frac{ 1}{3x^2 + 2} \bigg|_{x=1} = \frac{1}{5} \]

it then follows that the tangent line throughout point (2,1) is
\[ y - 1 = \frac{1}{5} (x-2) \\
y = \frac{1}{5} x +0.6 \]
see graph