Extreme values $f(x)=(x-2)^{\frac{1}{3}}$

Question

find all points of intrest for the function:

$f(x)=(x-2)^{\frac{1}{3}}$

Here we can clearly see that when $x=2$ $f(x)=0$ so I know that there atleast should exist a critical point.

Since the function is when taking it's derivative

I get

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

I know that this is an odd function therefor it will be symmetric about the origin and will alwaays be increasing. However accoriding to the formula there are no max or min values.

Which confuses me since singular points are found where the derivative is undefined i.e at $x=2$ so why is't the function either a max/min at $x=2$?

Many thanks to whomever might help me with this!

Answer 1

First, you said

Here we can clearly see that when $x=2$ $f(x)=0$ so I know that there at least should exist a critical point.

This is not a criteria for a critical point. A critical point occurs when $f'(x) = 0$

Second, you said

I know that this is an odd function

$f'(x)$ not not an odd function. You can tell just from the fact that it's only defined on $(2,\infty)$. The definition of an odd function is $f(-x) = -f(x)$. $x=0$ isn't anywhere in the domain, so $f(-x)$ doesn't even make any sense.

The reason we're considering $x=2$ is because it's on the boundary of the function's domain, which is $[2,\infty)$. Because $f'(x) > 0, \forall x > 2$ , $f(x)$ is increasing on the interval $(2,\infty)$. This makes $x=2$ a minimum of the function.

Singular points in the derivative have nothing to do with the function's max/min.

---

Addendum:

I know that this is an odd function therefor it will be symmetric about the origin and will always be increasing.

If we're being generous and interpret the function as $f(x) = \sqrt[3]{x-2}$, which is defined for $x < 2$, this still wouldn't make $f(x)$ an odd function. You can say that it's anti-symmetric about the point $x=2$ since $\sqrt[3]{2-x} = -\sqrt[3]{x-2}$.

Answer 2

Max and min occure when a derivative $f'(x)$ is equal to $0$. If the derivative is undefined then.... it is not equal to $0$.

If we think of the derivative as the "slope" of the tangent line of the function at a point $x$ then $f'(x)$ being undefined means the slope is infinite at that point. .... which is precise what happens and $x = 2$. Then function $f$ is at that instant increasing at an infinite rate.

(It only has infinite increasing at a single infinitisimal point so that is quite possible.)

....

Here we can clearly see that when $x=2$ $f(x)=0$ so I know that there atleast should exist a critical point.

That's not a critical point. That is an $x$-intercept. It's point of interest because ... it is a point where the graph intercepts that $x$ axis, that other than that it doesn't say anything else. It is not a critical point.

The other point of inters would be the $y$-intercept, where $x = 0$ and $f(x) = -(\frac 12)^{\frac 13}$. That is where the graph intercepts the $y$ axis. It is also not a critical point.

I know that this is an odd function therefor it will be symmetric about the origin

It is not an odd function $f(-x)=(-x-2)^{\frac{1}{3}}\ne -(x-2)^{\frac{2}{3}}=-f'(x)$. So it is not odd.

It is symmetric around the $x = 2$ line.

and will alwaays be increasing.

It'll be always increasing if $f'(x) > 0$ always.

Which in this case $f'(x) = \frac{1}{3(x-2)^{\frac{2}{3}}} = \frac 1{3\sqrt[3]{(x -2)^2}} > 0$ so $f$ is always increasing.

However accoriding to the formula there are no max or min values.

That's to be expected.

If something is always increasing (or decreasing) it can't have a max or a min.

To have a minimum the (continuous) function must decrease to a lowest point... and then stop decreasing and start increasing. Likewise to have a maximum the function must increase to a maximum. Stop increasing. And then decrease.

---

Here the image:

As you can see the function not symetric at the $y$-axis so it is not even nor is rotational around the origin so it is not odd. It is rotation around the $(2,0)$ point so it is "odd-like".

Now look at the point $(2,0)$. See how the function is going straight up and down. That is because the derivative $f'(x)=\frac{1}{3(x-2)^{\frac{2}{3}}}$ is undefined at $x=2$ and would evaluate to $\frac 10$ which is undefined. The function's tangent line has infinite slope. It goes straight up and down.

Now, notice there is no way any sane person using basic english language could call that a minimum (It gets immediately smaller when $x < 2$), or maximum (It gets immediately larger at $x > 2$), nor a saddle-point (you can't sit on it-- you'd fall off). I suppose we could call it a stripper-pole point but... well, thank goodness we don't.

Now in response to a comment here we have the function $g(x) = (x-2)^{\frac 23}x$.

Where $g(A) > g(c)$ and $g(B) > g(c)$ and $A < c < B$.

But notice $c$ is not the only such point that is true of nearly all the points between $A$ and $B$.

It has $g'(x) =\frac{1}{3(x-2)^{\frac{2}{3}}}x + (x-2)^{\frac 13}$

$g'(2)$ is still undefined and we still have "stripper-pole" point at $(2,0)$.

But at $x = 1 \frac 23$ we have $g'(x) = 0$. And at that point, a different point, we have a critical point. A minimum.

So $x < 1\frac 23$ we see $g'(x)< 0$ and $g$ is decreasing.

At $x = 1\frac 23$ we see $g'(x) = 0$ and we have "flatline". We've hit bottom. It's a local (and as it turns out global) minimum.

At $x > 1\frac 23$ we see $g'(x) > 0$ and the $g$ it increases. ... very fast.

At $x = 2$ it is increasing infinitely fast and the graph is straight up and down. $g'(2)$ is undefined and we have a "stripper-pole" point. These points, are increasing or decreasing infinitely fast so that can only exist at a single point.

At $x > 2$ we so that $g'(x) > 0$ and the function is still increasing, but not as fast. It hit a maximum rate of increase at $x = 2$.

Answer 3

You are interested in the function $$f(x)=(x-2)^{\frac{1}{3}}$$

This function is continuous and well defines on $(-\infty, \infty )$

It is symmetrical around the point $(2,0).$

The intercepts are $(0,-2^{1/3})$ and $(2,0)$

It is not differentiable at x=2 because its derivative is $ f'(x)=\frac{1}{3(x-2)^{\frac{2}{3}}}$

It is increasing on $(-\infty, \infty )$

It is one-to-one so it has an inverse function.

It is concave up for $ x<2$ and concave down for $x>2$