# Find a Cubic Function given an inflection point, critical point, and function value.

Find a cubic function f(x) = ax^3 + bx^2 + cx + d

Given:

• Inflection point (0,18)
• Critical point x = 2
• F(2) = 2

I know how to solve for the general forms of the derivatives, and to set the values of the functions and the derivatives at those points, but the system of equations that I come up with lead me to the wrong answer.

My work so far is as follows:

• f(x) = ax^3 + bx^2 + cx + d
• f'(x) = 3ax^2 + 2bx + c
• f''(x) = 6ax + 2b

• f(2) = a(2)^3 + b(2)^2 + c(2) + d = 2
• f(2) = 8a + 4b + 2c + d =2
• f'(2) = 3a(2)^2 + 2b(2) + c = 0
• f'(2) = 12a + 4b +c = 0
• f''(18) = 6a(18) + 2b = 0
• f''(18) = 108a + 2b = 0

Thus:

• f(2) = 8a + 4b + 2c + d =2
• f'(2) = 12a + 4b +c = 0
• f''(18) = 108a + 2b = 0

But here I get stuck and unable to solve for any particular variable.

• Your question would be better received if you include the system of equations you came up with. Moreover, we can check where in the equations did you get it wrong. – player3236 Oct 22 at 19:01
• Good point thank you for the comment. I added my work. – QBEE Oct 22 at 19:17
• As you have noticed, you simply mixed up the $x, y$ coordinates of the point of inflection. – player3236 Oct 22 at 19:21
• Yup my mistake has become apparent. I am glad this forum exists because I have been stumped for hours on this. Thank you for the help. – QBEE Oct 22 at 19:26

$$f''(0) = 0$$ and $$f(0)=18$$.
The critical point gives rise to the equation $$f'(2)=0$$ and you have $$f(2)=2$$. Then you have four linear equations in four unknowns, which you can solve by substitution.
• I meant of course $f'(2)=0$. – Fenris Oct 22 at 19:09