Calculating Derivatives and Finding Roots in Math

[Petrus]

Hello MHB,
I got one question, I was looking at a Swedish math video for draw graph and for some reason he did take derivate and did equal to zero and did calculate the roots and then he did take limit of the derivate function to the roots and it's there I did not understand, what does that mean? example we get the root 1 then we take the \(\displaystyle \lim_{x->1}f'(x)\) what does that mean?

Regards,
\(\displaystyle |\pi\rangle\)

 

[alyafey22]

Hello MHB,
I got one question, I was looking at a Swedish math video for drawing graphs and for some reason he took the derivative and did equal it to zero and calculated the roots and then he took limit of the derivative function as it approaches the roots it's there I did not understand, what does that mean? .For example we get the root 1 then we take the \(\displaystyle \lim_{x->1}f'(x)\) what does that mean?

Regards,
\(\displaystyle |\pi\rangle\)

Let us take an example , and see what happens . For simplicity , I will choose a polynomial so the limit always exists .

\(\displaystyle P(x) = x^3+2x^2+1\) .

Now , make the procedures and write the full answer , tell me what you think .

 

[Petrus]

Let us take an example , and see what happens . For simplicity , I will choose a polynomial so the limit always exists .

\(\displaystyle P(x) = x^3+2x^2+1\) .

Now , make the procedures and write the full answer , tell me what you think .

Derivate it and equal to zero we get the roots.
\(\displaystyle x_1=0 \ x_2= -\frac{4}{3}\)
\(\displaystyle \lim_{x->0} 3x^2+4x= 0\)
\(\displaystyle \lim_{x->-\frac{4}{3}} 3x^2+4x= 0\)
Does that mean the slope is zero at those x point?

Regards,
\(\displaystyle |\pi\rangle\)

 

[alyafey22]

Derivate it and equal to zero we get the roots.
\(\displaystyle x_1=0 \ x_2= -\frac{4}{3}\)
\(\displaystyle \lim_{x->0} 3x^2+4x= 0\)
\(\displaystyle \lim_{x->-\frac{4}{3}} 3x^2+4x= 0\)
Does that mean the slope is zero at those x point?

Regards,
\(\displaystyle |\pi\rangle\)

We will see !.Let us take an interesting question \(\displaystyle f(x) = \sqrt{x^2-x^3}\)

 

[Petrus]

We will see !.Let us take an interesting question \(\displaystyle f(x) = \sqrt{x^2-x^3}\)

roots we get is \(\displaystyle \frac{2}{3}\), zero is also a root but we don't accept it cause the bottom will get also zero right?
and the limit becomes zero as well here.

Regards,
\(\displaystyle |\pi\rangle\)

 

[Evgeny.Makarov]

roots we get is \(\displaystyle \frac{2}{3}\), zero is also a root but we don't accept it cause the bottom will get also zero right?

Yes, both 2/3 and 0 are critical points: 2/3 because the derivative is 0 and 0 because the derivative does not exist.

and the limit becomes zero as well here.

Which limit? Neither $\lim_{x\to0^-}f'(x)$ nor $\lim_{x\to0^+}f'(x)$ is zero.

When the derivative is continuous (i.e., the function is of class C¹) and the derivative is 0 at some point, then there is no sense in taking the limit of the derivative. This in particular happens when the derivative is expressed using usual functions (arithmetical operations, roots, trigonometric functions, which are all continuous on their domains) and is defined in some neighborhood of the critical point. Taking limits of the derivative makes sense when the derivative does not exist in a critical point, as with the function above.