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

##### 1 Answer

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:

Now,

And of course

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

=

=

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:

=

=

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

So

Rewrite this as:

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

=

or

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