more on maxima and minima

[logistic_guy]

Find the critical numbers of \(\displaystyle f(x) = \frac{x^2}{x - 1}\), and determine whether they yield relative maxima, relative minima, or inflection points.

 

[khansaheb]

Find the critical numbers of \(\displaystyle f(x) = \frac{x^2}{x - 1}\), and determine whether they yield relative maxima, relative minima, or inflection points.

What is the definition of "critical point" in the above context?

 

[logistic_guy]

Let us take the first derivative of the function.

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

Next, we set what we got to zero.

\(\displaystyle 0 = \frac{x^2 - 2x}{(x - 1)^2}\)

\(\displaystyle 0 = x^2 - 2x\)

\(\displaystyle 0 = x(x - 2)\)

This gives:

\(\displaystyle x = 0\) or \(\displaystyle x = 2\)

We also want to know when the first derivative is undefined, so we have:

\(\displaystyle 0 = (x - 1)^2\)

\(\displaystyle 0 = x - 1\)

\(\displaystyle x = 1\)

Since \(\displaystyle x = 1\) is not in the domain of the function \(\displaystyle f\), we ignore it.

Therefore, we have found two critical numbers from the first derivative.
\(\displaystyle x = 0\)
\(\displaystyle x = 2\)

In the next post, we will take the second derivative and we will see if there are more critical numbers available there.

 

[logistic_guy]

Let us take the second derivative.

\(\displaystyle f''(x) = \frac{(2x - 2)(x - 1)^2 - (x^2 - 2x)2(x - 1)}{(x - 1)^4} = \frac{(2x - 2)(x - 1) - (x^2 - 2x)2}{(x - 1)^3}\)


\(\displaystyle = \frac{2x^2 - 2x - 2x + 2 - 2x^2 + 4x}{(x - 1)^3} = \frac{2}{(x - 1)^3}\)

Since we cannot sit this to zero, we know that there are no inflection points. Let us also check for critical numbers where the second derivative is undefined.

\(\displaystyle (x - 1)^3 = 0\)

\(\displaystyle x - 1 = 0\)

\(\displaystyle x = 1\)

We ignore \(\displaystyle x = 1\) since it is not in the domain of the function \(\displaystyle f(x)\).

So, all the critical numbers that we have found are:
\(\displaystyle x = 0\)
\(\displaystyle x = 2\)

In the next post, we will test those critical numbers.

 

[khansaheb]

What is the definition of a "critical point" in the above context?

 

[logistic_guy]

What is the definition of "critical point" in the above context?

My solution speaks for itself.

If we substitute the critical numbers in the second derivative, we will have a pretty good information about their maxima or minima.

Let us start with \(\displaystyle x = 0\).

\(\displaystyle f''(0) = \frac{2}{(0 - 1)^3} = -2\)

This tells us that the critical number \(\displaystyle x = 0\) yields a relative maxima.

 

[logistic_guy]

Now let us check \(\displaystyle x = 2\).

\(\displaystyle f''(2) = \frac{2}{(2-1)^{3}} = 2\)

This means that \(\displaystyle x = 2\) yields a relative minima.