The local min is $(3,3)$ and the local max is $(5,1)$ with an inflection point at $(4,2)$
The general formula of a cubic function
$$f(x)=ax^3+bx^2+cx+d $$
The derivative of which is
$$f'(x)= 3ax^2+2bx+c $$
Using the local max I can plug in $f(1)$ to get
$$f(1)=125a+25b+ 5c+d $$
The same goes for the local min
$$f(3)=27a+9b+3c+d$$
But where do I go from here? Should I subtract the two equations from one another or use the deriviative to find some constant?
Best Answer
$f'(x)=A(x-3)(x-5)=A(x^2-8x+15)$ for some constant $A$.
$f(x)=A\left(\dfrac{x^3}3-4x^2+15x\right)+k$ for some constant $k$.
Since $f(3)=3$ and $f(5)=1$, $3=18A+k$ and $1=\dfrac{50}3A+k$.
Solving, $A=\dfrac32$ and $k=-24$.
$f(x)=\dfrac12x^3-6x^2+\dfrac{45}2x-24$.
Note that $f''(x)=3x-12$.
$f''(4)=0$ and $f(4)=2$.