Question

Is `f(x) =(x-2)^2/(x-1)` concave or convex at `x=2`?

Answer 1

Concave up

Explanation:

You have to find the 2nd derivative to determine this.
So, take the derivative, then do it again.

The first derivative is the derivative of the quotient of 2 functions:

`d/dx f(x)/g(x) = (f'(x)g(x) - f(x) g'(x))/(g(x)^2)`

Now, `d/dx (x-2)^2` = `d/dx (x^2 - 4x + 4)` = 2x - 4

And of course `d/dx (x-1)` = 1

so, fitting these into our forumula for the derivative of the quotient of two functions, we have

`d/dx f(x)/g(x) =((2x -4)(x - 1) - (x^2 - 4x + 4))/(x - 1)^2`

= `((2x^2 - 6x + 4) - (x^2 - 4x + 4))/(x - 1)^2`

= `(x^2 - 2x)/(x-1)^2`

We can evaluate this at x = 2. The value is 0, so x = 2 is indeed a critical point. (Minima or maxima).

To find if it's a minima (function is concave up at this point) or maxima (concave down), we have to take the derivative again.

I used a little algebraic sleight of hand to simplify taking the second deriviate:

`(x^2 - 2x)/(x-1)^2 = (x^2 - 2x + 1 - 1)/(x-1)^2`

= `(x-1)^2/(x-1)^2 - 1/(x-1)^2`

= `1 - 1/(x-1)^2`

To find the derivative of this, note that the constant 1 in this equation drops out.

So `d/dx (1 - 1/(x-1)^2)` = `d/dx (- 1/(x-1)^2)`

Rewrite this as:

`d/dx (-(x-1)^-2)`

which is an application of the chain rule you should be able to do in your head:

= `2(x-1)^-3`

or `2/(x-1)^3`

At a value of x=2, this evaluates to 2. Which, being > 0, indicates the curve is concave upward at this point.

You can double check this at:
https://www.desmos.com/calculator

...paste in:
\frac{\left(x\ -2\right)^2}{x-1}

so see the graph of the original function