Are There Global Extrema for These Functions?

[mathmari]

Hey!

I have show that the function $y=f(x)=x-\frac{5}{2}+\frac{4}{x}=\frac{2x^2-5x+8}{2x}$ has a local minimum at $x=22$ and a local maximum at $x=-2$. How ca we check if they are also global extrema? (Wondering)

The same at the function $f(x)=10e^{-x}(x-1)^2$, at $x=1$ the function has a local minimum and at $x=3$ it has a local maximum. What about the global minimum/maximum? (Wondering)

 

[I like Serena]

Hey!

I have show that the function $y=f(x)=x-\frac{5}{2}+\frac{4}{x}=\frac{2x^2-5x+8}{2x}$ has a local minimum at $x=22$ and a local maximum at $x=-2$. How ca we check if they are also global extrema? (Wondering)

The same at the function $f(x)=10e^{-x}(x-1)^2$, at $x=1$ the function has a local minimum and at $x=3$ it has a local maximum. What about the global minimum/maximum? (Wondering)

Hi mathmari! (Smile)

Since those functions are continuously differentiable, it suffices to check what happens on the boundaries of the domain.
That's because any extremum at an interior point of the domain has a derivative of $0$.
It's only at the boundaries where the function can have an extremum that does not necessarily have a derivative of $0$.

For the first function that means checking the values of:
$$\lim_{x\to -\infty} f(x), \quad
\lim_{x\uparrow 0} f(x), \quad
\lim_{x\downarrow 0} f(x), \quad
\lim_{x\to +\infty} f(x)
$$
(Thinking)

 

[mathmari]

Hi mathmari! (Smile)

Since those functions are continuously differentiable, it suffices to check what happens on the boundaries of the domain.
That's because any extremum at an interior point of the domain has a derivative of $0$.
It's only at the boundaries where the function can have an extremum that does not necessarily have a the derivative of $0$.

For the first function that means checking the values of:
$$\lim_{x\to -\infty} f(x), \quad
\lim_{x\uparrow 0} f(x), \quad
\lim_{x\downarrow 0} f(x), \quad
\lim_{x\to +\infty} f(x)
$$
(Thinking)

We have that $$\lim_{x\to -\infty} f(x)=-\infty, \quad
\lim_{x\uparrow 0} f(x)=+\infty, \quad
\lim_{x\downarrow 0} f(x)=-\infty, \quad
\lim_{x\to +\infty} f(x)=+\infty
$$ right? (Wondering)

What do we conclude from that? (Wondering)

 

[I like Serena]

We have that $$\lim_{x\to -\infty} f(x)=-\infty, \quad
\lim_{x\uparrow 0} f(x)=+\infty, \quad
\lim_{x\downarrow 0} f(x)=-\infty, \quad
\lim_{x\to +\infty} f(x)=+\infty
$$ right? (Wondering)

What do we conclude from that? (Wondering)

Right!
We conclude that there is no global maximum nor global minimum. (Wink)

 

[mathmari]

For the other function we have $$\lim_{x\to -\infty} f(x)=\lim_{x\to -\infty} 10e^{-x}(x^2-2x+1)=+\infty, \quad
\lim_{x\to +\infty} f(x)=\lim_{x\to +\infty} \frac{10(x^2-2x+1)}{e^x}=0$$

So, does this mean that there is a global extrema? Suppose we have also the function $g(x)=\frac{x}{x^2+1}$, then $$\lim_{x\to -\infty} g(x)=-\infty, \quad
\lim_{x\to +\infty} g(x)=+\infty$$ then we conclude that there is no global extrema, right? (Wondering)

 

[I like Serena]

For the other function we have $$\lim_{x\to -\infty} f(x)=\lim_{x\to -\infty} 10e^{-x}(x^2-2x+1)=+\infty, \quad
\lim_{x\to +\infty} f(x)=\lim_{x\to +\infty} \frac{10(x^2-2x+1)}{e^x}=0$$

So, does this mean that there is a global extrema?

It means there is no global maximum, but there might be a global minimum.
And we know now that we we have one horizontal asymptote, which is $y=0$. (Nerd)

Suppose we have also the function $g(x)=\frac{x}{x^2+1}$, then $$\lim_{x\to -\infty} g(x)=-\infty, \quad
\lim_{x\to +\infty} g(x)=+\infty$$ then we conclude that there is no global extrema, right? (Wondering)

Yes. The function takes on values that are higher and lower than any possible maximum respectively minimum could be.

 

[mathmari]

Yes. The function takes on values that are higher and lower than any possible maximum respectively minimum could be.

I computed wrong the limits at $g$. It must be $\lim_{x\rightarrow -\infty}g(x)= \lim_{x\rightarrow +\infty}g(x)=0$, right? (Wondering)
Haveing found the local extrema, do we conclude that they must be then also global? (Wondering)

 

[I like Serena]

I computed wrong the limits at $g$. It must be $\lim_{x\rightarrow -\infty}g(x)= \lim_{x\rightarrow +\infty}g(x)=0$, right?
Haveing found the local extrema, do we conclude that they must be then also global?

We have to inspect them to tell.
In this case we have one local extremum that is higher than $0$, meaning it's the global maximum.
And we have one local extremum that is lower than $0$, meaning it's the global mimimum. (Nerd)

 

[mathmari]

I understand! (Nerd)

A function has an inflection point only when the second derivative has a root, right? (Wondering)

 

[I like Serena]

A function has an inflection point only when the second derivative has a root, right?

Close, although it's a little more subtle as we can see in wiki:

Inflection points are the points of the curve where the curvature changes its sign while a tangent exists.
[...]
For a twice differentiable function, an inflection point is a point on the graph at which the second derivative has an isolated zero and changes sign.​

So for instance when the function is not twice differentiable, I think we can still have an inflection point. (Nerd)

 

[mathmari]

I see... Thanks a lot! (Happy)