Let $f'(x)=\dfrac{(x-1)(x-8)(x-6)^2}{(x-2)(x-9)^3(x-7)^2}$. Then find point of extrema and point of inflection of $f(x)$

Question

Let $f'(x)=\dfrac{(x-1)(x-8)(x-6)^2}{(x-2)(x-9)^3(x-7)^4}$. Then find point of extrema and point of inflection of $f(x)$ whose derivative is given function.

Solution by my Teacher:

He said $x=\{2,9\}$ is point of minima and minima is lost and $\{1, 8\}$ are point of maxima. I understood this statement.

Then He said $x=7$ is point of inflection but point of inflection is lost. I don't understand why point of inflection is lost. For inflection point only requirement is concavity should change. And in this question concavity is changing.

Then He took example of $\tan x$ and said $x=\dfrac{\pi}{2}$ is not point of inflection because function must be continuous for point of inflection. Is He correct about this statement?

Also He said in above question we will not have vertical tangent at $x=7$.

Later He said if question was $f'(x)=\dfrac{(x-1)(x-8)(x-6)^2}{(x-2)(x-9)^3(x-7)^2}$ then $x=7$ will have vertical tangent.

Can anyone out here explain me everything specillay when denominator is containing $(x-7)^4$ and $(x-7)^2$.

Answer 1

Try with simpler function. Can you say about the inflection point for $f(x)=x^3$? What can you say for second derivative test?

Can you guess how the multiplicity related with the Vertical Asymptotes?

Let $x=c$ be a Vertical Asymptote of a rational function with multiplicity $m$. i) If $m$ is even, then the sign of function does NOT change from one side to the other side of $c$, or the graph touches but doesn't cross $\infty$ at c. ii) If $m$ is odd, then the sign of the function DOES change from one side to the other side of $c$, or the graph crosses the $\infty$ at $c$.