So I need to find a formula for a cubic polynomial with the given conditions:
- Local max at $x=1$
- Local min at $x=3$
- $Y$ intercept of $5$
- $x^3$ term whose coefficient is $1$
So all I really have is $x^3+x^2+x+5$.
I know that $F'(1) = 0$ and $F''(1) < 0$. Also $F'(3) = 0$ and $F''(3) > 0$. But as far as setting up the problem, I'm lost.
$\endgroup$ 13 Answers
$\begingroup$You don't have $f(x)=x^3+x^2+x+5$; you have $f(x)=x^3+bx^2+cx+5$ where the coefficients $b$ and $c$ are yet unknown.
Since you also have $f'(1)=0$ and $f'(3)=0$, this gives you two (linear) equations in the two unknowns $b$ and $c$ which you can solve.
This doesn't use the information that $1$ and $3$ are maximum and minimum rather than the other way around, but for a third-degree polynomial whose leading coefficient is positive, the local maximum (if any) will always come before the local minimum (if any), so that is automatically satisfied.
$\endgroup$ $\begingroup$Clearly the first derivative of a cubic polynomial will be a quadratic, suppose $g(x)$.
According to the question, the function has local maxima and local minima at $1$ and $3$ respectively. Hence we have $f'(x)=g(x)=0$ at $x=1,3$. Therefore $g(x)$ can be written as $$\begin{align} g(x)&=A(x-1)(x-3)\\ g(x)&=A(x^2-4x+3)\\ f(x)&=A\int (x^2-4x+3~dx)\\ \end{align}$$
Finally apply the condition $f(0)=5$ to get the constant of integration and use the fact that the coefficient of $x^3$ has to be 1 to get A.
$\endgroup$ 2 $\begingroup$You need to add 2 coefficients to your equation:
${x^3} + a{x^2} + bx + 5$
Then I think you should be able to solve for a & b using only the 1st derivatives.
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