chaomeinguyzyp3

2023-02-27

What are the global and local extrema of $f\left(x\right)={x}^{3}-{x}^{2}-x+1$ ?

Addison Stuart

Beginner2023-02-28Added 5 answers

Local extrema, also called maxima & minima, or sometimes critical points, are just what they sound like: when the function's reached a brief maximum or a brief minimum. They are called local because, when searching for crucial points, you frequently are only interested in what the maximum means in the point's immediate vicinity.

Finding local critical points are pretty simple. Find when the function is unchanging, and the function is unchanging when - you guessed it - the derivative equals zero.

A simple application of the power rule gives us $f\prime \left(x\right)$,

$f\prime \left(x\right)=3{x}^{2}-2x-1$.

We're concerned when this expression equals zero:

$0=3{x}^{2}-2x-1$

Now we've found ourselves looking at a quadratic equation in $x$, which should be fairly easy to solve.

There are indeed two real-valued solutions to this quadratic, given by the quadratic formula or your method of choice, and they are $x=-\frac{1}{3}$ and $x=1$.

So we've determined that there are two local extrema, as well as their locations. Classifying whether each point is a maximum or minimum is a different story and I won't go into that here, but I can direct you here if that's something you'd like to read up on.

Now, on to the global extrema. A global extremum is defined as being the single maximum or single minimum point of a function on a whole interval. Usually, the interval is given, such as ""find the global extrema of such-and-such on the interval $[0,3]$,"" but it can also be the entire domain of the function.

With global extrema, there's more you need to take into account than just the derivative. You would have to determine if there are any critical points on this interval, because if so, one might (but not necessarily) be the global extrema as well. With these types of situations, having a calculator plot is the most helpful, but a little analysis reveals the critical points. (I can direct you to this page for more info and a few examples)

In this case, the function continues getting really, really huge as $x\to \infty$, and gets closer to $-\infty$ as $x\to -\infty$. So, there really is no global maximum or minimum - there are only the two local critical points.

Finding local critical points are pretty simple. Find when the function is unchanging, and the function is unchanging when - you guessed it - the derivative equals zero.

A simple application of the power rule gives us $f\prime \left(x\right)$,

$f\prime \left(x\right)=3{x}^{2}-2x-1$.

We're concerned when this expression equals zero:

$0=3{x}^{2}-2x-1$

Now we've found ourselves looking at a quadratic equation in $x$, which should be fairly easy to solve.

There are indeed two real-valued solutions to this quadratic, given by the quadratic formula or your method of choice, and they are $x=-\frac{1}{3}$ and $x=1$.

So we've determined that there are two local extrema, as well as their locations. Classifying whether each point is a maximum or minimum is a different story and I won't go into that here, but I can direct you here if that's something you'd like to read up on.

Now, on to the global extrema. A global extremum is defined as being the single maximum or single minimum point of a function on a whole interval. Usually, the interval is given, such as ""find the global extrema of such-and-such on the interval $[0,3]$,"" but it can also be the entire domain of the function.

With global extrema, there's more you need to take into account than just the derivative. You would have to determine if there are any critical points on this interval, because if so, one might (but not necessarily) be the global extrema as well. With these types of situations, having a calculator plot is the most helpful, but a little analysis reveals the critical points. (I can direct you to this page for more info and a few examples)

In this case, the function continues getting really, really huge as $x\to \infty$, and gets closer to $-\infty$ as $x\to -\infty$. So, there really is no global maximum or minimum - there are only the two local critical points.