The point of inflection on the curve $=^3−^2−+$ is a stationary point of inflexion. Show that $b=8a^2$. The point of inflection on the curve $=^3−^2−+$ is a stationary point of inflection. Show that $b=8a2$. Thank you for your help. Edit: The solution to this question drafted in another post is wrong, according to the book. That is why I post it again, in a separate post. Edit 2: I am not sure, then, why some of you still mark it as a duplicate.

To find stationary points we need $f'(x)=0$ and to find inflection points we need to find where $f''(x)=0$.

$$f(x)=x^3-ax^2-bx+c$$ $$f'(x)=3x^2-2ax-b$$ $$f''(x)=6x-2a$$

So to find the inflection point we need $$6x-2a=0$$ $$6x=2a$$ $$x=\frac{a}{3}.$$

In order for this point to be stationary we need $f'(\frac{a}{3})=0$.

$$3\bigg(\frac{a}{3}\bigg)^2-2a\bigg(\frac{a}{3}\bigg)-b=0$$ $$\frac{a^2}{3}-\frac{2a^2}{3}-b=0$$

Now you should be able to plug in $a$ and $b$ and show that this equation is correct. Is your question saying that $a=2$ and $b=8$? Or $b=8a^2$? In either case, this inflection point would not be a stationary point in this function.