I was reading what inflection and critical points are here. If we assume that a function is defined on an open interval. Then does it mean we can say that all inflection points of the function are critical points of the function?
Inflection points vs critical points
calculusderivativesreal-analysis
Related Solutions
Yes, you find inflection points by taking the second derivative $y''$ and setting $y''$ equal to zero. Solve for x, to determine the point $(x, y)$ at which an inflection point may occur. (This procedure may not result in an inflection point, but in this case it does. If an inflection point exists, it will be at the point at which $y'' = 0$, but the converse is not always true.
For your critical points, find the corresponding $y$ values associated with your solutions fr $x$, respectively, to obtain the critical points: you want to write the points as ordered pairs.
The derivative has to be zero at 3 points, and at one of them, it has to have a double zero to get that saddle point. This is a total of 4 zeros for the derivative, meaning it has to be a fourth order polynomial where the zeros are at known locations, meaning I can write it as:
$$ f'(x) = (x-x_1)(x-x_2)(x-x_3)^2, $$ where $x_1$ is the max point, $x_2$ is the min point, and $x_3$ is the saddle point. You can plug in some some concrete values and get a form for $f'$, then integrate. So the most general formula for your curve is
$$ f(x) = \int(x-x_1)(x-x_2)(x-x_3)^2\;dx $$
I used some concrete values ($x_1=1/8,\;x_2=1/2,\;x_3=1$) and got
$$ f(x) = 1/16 x - 3/8 x^2 + 37/48 x^3 - 21/32 x^4 + x^5/5 $$
Take a look here so you can put in your own values if you want and get the polynomial coefficients for LaTeX use.
EDIT If you really want that the function must be zero and have zero derivative at $x_2$, then you can begin as mentioned in the comments, and add in an offset:
$$ g(x) = f(x) - f(1/2) $$
So clearly $g(1/2)=0$, but now $g(0)\neq 0$. However, we can just move the curve to the right by the correct amount now to get the curve to go through the origin. This requires finding where $g(x_4)=0$ (a small shift in this problem of about $x_4=-0.01346503$, this has a closed form in terms of radicals), but once we have that, we form a shifted function
$$ h(x) = g(x+x_4). $$
The function $h$ satisfies all the requirements, as you can see through the links below. If you want the final, exact coefficients, they are really messy (here). You can also get approximate coefficients using the decimal approximation above. With the approximation, $h(0)\approx 10^{-10}$ which isn't exactly zero, but it is graphically indistinguishable from zero for your plot.
Best Answer
No.
Critical points of a function are where a function has a horizontal or vertical tangent, or is at a defined point where the function is not differentiable.
Points of inflection are where a function changes its concavity.
For example, take the function $f(x)=x^3-12x$. Its derivative is $f'(x)=3x^2-12$ and its second derivative is $f''(x)=6x$. It has two critical points at $(-2,16)$ and $(2,-16)$, and a single point of inflection at $(0,0)$.