Inverse functions (1 Viewer)

[ashokkumar]

Let y = x^3 - 3x^2 - 1 (for x >=2). Find the slope of the tangent to the inverse function at x = -1.

 

[Trebla]

Let y = x^3 - 3x^2 - 1 (for x >=2). Find the slope of the tangent to the inverse function at x = -1.

Let f(x) = x³ - 3x² - 1
Let the inverse for x ≥ 2 be y = f^-1(x), we seek dy/dx at x = - 1
If y = f^-1(x), then
x = f(y)
dx/dy = f'(y)
dy/dx = 1 / f'(y)

At x = - 1, y = f^-1(-1) = a (which we need to find)
So the point of contact of the tangent on the inverse function is (-1, a) but the reflective point on the original function is (a, -1) which implies f(a) = - 1
=> a³ - 3a² - 1 = - 1
a²(a - 3) = 0
a = 0, 3
But since the function is defined for x ≥ 2 we choose a = 3 hence the point on the inverse function is (-1, 3) so
dy/dx = 1 / f'(3) at x = -1

Since f(x) = x³ - 3x² - 1
f'(x) = 3x² - 6x
f'(3) = 9
=> dy/dx = 1/9

 

[ashokkumar]

is there a simpler way of doing this?

I simply differentiated the the original equation, found the reciprocal (i.e. dx/dy) and then subbed in x = -1 and i got the same answer (1/9).

Or is that method incorrect?

 

[Trebla]

is there a simpler way of doing this?

I simply differentiated the the original equation, found the reciprocal (i.e. dx/dy) and then subbed in x = -1 and i got the same answer (1/9).

Or is that method incorrect?

I'm not sure about the logic behind that...I think the way you got the same answer might be a coincidence...

If y = f(x), then
x = f^-1(y)
dy/dx = f'(x)
dx/dy = 1 / f'(x)
=> d[f^-1(y)]/dy = 1 / f'(x)
x = -1 refers to the input value of the inverse. In the way it was defined, the input value of the inverse function is y from your approach, using x = f^-1(y) so really you're subbing in y = -1 and solving it the same way (just x and y are the other way around)