# Does tangent line of inflection point always passes through the curve?

I saw many functions on my book and all of the tangent line of inflection point always pass through the curve, Here are examples :

Example 1 :

$$f(x) = x^3 \quad (x=0)$$

Tangent line at $$x = 0$$, $$l:y=0$$

Then we know that $$l$$ passes through the $$f(x)$$.

Example 2 : $$f(x)=\sin x\quad (x=0)$$

Tangent line at $$x = 0$$, $$l: y = x$$

We know $$l$$ passes through the $$f(x)$$.

Let $$f$$ is differentiable function and $$f$$ is not a constant, linear function.

If a line $$l$$ is a tangent line of inflection point, $$l$$ passes through $$f$$ near of inflection point?

I think this is true, but I have no idea to prove this. Thanks for help.

Yes. Some people define inflection point as a point where the tangent line crosses the curve. But most people define it as a point where the curve changes curvature.

If you think about the tangent line at a moving point, if the curve is concave down, the tangent line rotates clockwise as you move from left to right, and therefore it is above the curve. If the curve is concave up, the tangent line is rotates counter-clockwise and therefore is below the curve. So at the inflection point, the tangent line has to change sides of the curve.

Another way to show this is

Let $$l(x)$$ to be the tangent line of $$f(x)$$ at $$x=a$$.

Case-$$1$$: $$f''(xa)$$, $$f'(xa)$$

$$x\,\, f(a)+\int_a^xf'(t)\,dt=f(x)$$ $$x>a:\quad l(x)=f(a)+\int_a^xf'(a)\,dt \,\,<\,\, f(a)+\int_a^xf'(t)\,dt=f(x)$$

Case-$$2$$: $$f''(x0>f''(x>a)$$, $$f'(x f'(a) > f'(x>a)$$

$$x $$x>a:\quad l(x)=f(a)+\int_a^xf'(a)\,dt \,\,>\,\, f(a)+\int_a^xf'(t)\,dt=f(x)$$

So $$l(x)$$ passes through $$f(x)$$.

A tangent line always passes through the curve at a double or tangent point. The tangent line of inflection point always passes through the curve always, the but the center of curvature changes side onto the other side of the tangent.