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

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!

• Using Mathjax is a lot easier than using html. – suomynonA Mar 3 '17 at 4:00
• This is my first question so I barely knew what to use but I am more confusded about the answer though Thanks a lot for the edit though!BTW can you answer it? – Agent Smith Mar 3 '17 at 4:01
• Read through the link in the above comment so next time you can properly format your answer. – suomynonA Mar 3 '17 at 4:03
• sure I have test tommorow so I just didn't bother but I'll make sure next time – Agent Smith Mar 3 '17 at 4:04
• but you bothered to write <sup><\sup> which takes up more time? – suomynonA Mar 3 '17 at 4:04