A cubic polynomial with a local maximum at x = 1, a local minimum at x = 3, a y-intercept of 6, and an x3 term whose coefficient is 1.
ok, so how the flip do i do this bad johnny...thanks for any help..
mathaction
Ok...i assume you want to find the cubic.
Let the general form of the cubic be $\displaystyle f(x) = x^3 + ax^2 + bx + c$
we know that $\displaystyle c = 6$ since this is the y-intercept.
so we have: $\displaystyle f(x) = x^3 + ax^2 + bx + 6$
since the local max is at x = 1, it means $\displaystyle f'(1) = 0$ and $\displaystyle f''(1) < 0$
since the local min is at x = 3, it means $\displaystyle f'(3) = 0$ and $\displaystyle f''(3) > 0$
can you continue? (you may not need the second derivative, i just gave you extra information for your general edification)