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Graphing the first and second derivative of a 3rd degree polynomial function

Given the graph of h(x) (Attached Below), sketch the graphs of h'(x) and h''(x).

Please correct me if I am mistaken but I believe given the graph of h(x), the graph of the first derivative would be a parabola (opening up) with a minimum point at the coordinates (0,2) because there is a point of inflection on h(x) at x=2 and it is a continuously increasing function in the presented domain.

But is there anything other than that we can deduce about the graph of the first derivative given h(x)? Like the y-intercept for example? How accurate can my sketch be?

Thanks in advance.

Sincerely,

Raymond MacNeil

Re: Graphing the first and second derivative of a 3rd degree polynomial function

If it's the case that y = h(x) is a cubic, passes through the origin and the point with

co-ordinates (3,3), and has an inflection at x = 2, (as appears to be from the graph), then you can calculate its equation exactly, (in which case you can then calculate its derivatives exactly). If the co-ordinates of any of those points is in doubt, then you can't.

A good starting point is to assume that $\displaystyle y' =k(x-2)^2=k(x^2-4x+4),$ which gives you an inflection at$\displaystyle x=2.$

Re: Graphing the first and second derivative of a 3rd degree polynomial function

Quote:

Originally Posted by

**BobP** If it's the case that y = h(x) is a cubic, passes through the origin and the point with

co-ordinates (3,3), and has an inflection at x = 2, (as appears to be from the graph), then you can calculate its equation exactly, (in which case you can then calculate its derivatives exactly). If the co-ordinates of any of those points is in doubt, then you can't.

A good starting point is to assume that $\displaystyle y' =k(x-2)^2=k(x^2-4x+4),$ which gives you an inflection at$\displaystyle x=2.$

Yes, those points are correct. None are in doubt. The graph I made (which is attached) was a makeshift I created in MS Paint because the PDF File I'm working through is secured. (Couldn't copy and paste.) Thanks for this info. I will follow up on this later in more detail as I have to go now. Thanks for your reply. Perhaps later, I may ask for additional advice should I have any complications determining the equation of the cubic function.

All the best,

Raymond