Calculus Problem : Slope of the tangent line!!!

[dth257]

Let y=mx+b be the equation of the tangent line to the graph of y = f(x) at the point (x0,y0). What is the slope of the tangent line to graph of the inverse of f at the point (y0,x0)?
a/ m
b/ 1/m
c/ -m
d/ -1/m
e/ b
Please Help!!

 

[tkhunny]

You should have a Theorem, given appropriate conditions (which you should review): \(\displaystyle \dfrac{d}{dx}f^{-1}(x) = \dfrac{1}{f'(f^{-1}(x))}\)

 

[MarkFL]

\(\displaystyle y=f(x)\) implies:

\(\displaystyle x=f^{-1}(y)\)

and so:

\(\displaystyle \dfrac{dy}{dx}=f'(x)\)

\(\displaystyle \dfrac{dx}{dy}=(f^{-1}(y))'\)

Multiplying these equations, we find:

\(\displaystyle 1=f'(x)(f^{-1}(y))'\)

So, we must have:

\(\displaystyle 1=f'(x_0)(f^{-1}(y_0))'\)

So, what can we conclude then if \(\displaystyle f'(x_0)=m\)?

 

[dsk2]

dth257,

To find the inverse of the equation y=mx+c , simply swap x and y and rearrange the equation. In the original equation, the slope is 'm'.

The inverse equation is: x=my+c. Now, rearranging,

y=(x-c)/m=(1/m)x-(c/m)

Now, you know the slope of the inverse equation in terms of m.

Note: m is a constant that can be evaluated in terms of x0 and y0.

Cheers,
Sai.