Find degree function with max at $(2,44)$ and min at $(4,36)$

Question

A function has a local maximum at $(2,44)$ and a local minimum value at $(4,36)$. Find a degree function that has these qualities.

I know that the two important points for the $f (x)$ are $2$ and $4$ so $f'(x)=(x-2)(x-4)=x^2-6x+8$ so $f(x)$ should be equal to $f(x)=\dfrac{x^3}3 -3x^2 +8x+d$ but now when I try to find the value of d and it doesn't satisfy the points the answer in the textbook answer key says the answer may vary and that's all they said no example equation so I want to know is this function possible or not and if it is possible I want an example of it with a little bit of explanation of where I was wrong?

I got the answer for anyone who is wondering the answer is that as $f(x)=a[\dfrac{x^3}3 -3x^2 +8x]+d$ you can find values of a and d and get your equation now!

Answer 1

The question is a bit confusing. It was not clear if it is a single variable function or dual-variable function. As the points $(2,44)$ and $(4,36)$ are in the form of $(x,y)$, I consider $y=f(x)$ as the form. Thus, it is a function in one variable.

Now that $(2,44)$ and $(4,36)$ are two points which the function satisfies:

1. $f'(x=2)=0$ because $(2,44)$ is a local maxima. And, $f'(x=4)=0$ because $(4,36)$ is a local minima.
2. $f(x=2)=44$ and $f(x=4)=36$.

From condition 1, $$f'(x) = a (x-2)(x-4) \text{ where } a \ne 0$$ $$ \Rightarrow f(x) = a \bigg( \frac{x^3}{3}-3x^2+8x \bigg)+d$$

From condition 2, $$f(x=2)=44 \Rightarrow \frac{20}{3}a+b=44$$ and $$f(x=4)=36 \Rightarrow \frac{16}{3}a+b=36$$

Together they yield $$a=6 \text{ and } b=4$$ Thus, the function is $2x^3-18x^2+48x+4$