Hey guys,

Just wondering if someone can please answer this question! Could really use the explanation guys.

Cheers!

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- Dec 15th 2009, 09:18 AM #1

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- Dec 16th 2009, 02:40 AM #2

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The first and second derivatives require

**continuous**functions. Just individual values is not sufficient. I presume that you are really asked to**estimate**or**approximate**the derivatives.

My first thought would be to assume a linear function but then the second derivative would be 0, automatically. The next simplest thing to do is to assume a quadratic function. I would be inclined to write the function as $\displaystyle a(x-2)^2+ b(x-2)+ c$ so when x= 0, this is just $\displaystyle a(-2)^2+ b(-2)+ c= 4a- 2b+ c= 7$, when x= 2, this is just $\displaystyle a(0)^2+ b(0)+ c= c= 13$, and when x= 4, this is just [tex]a(2)^2+ b(2)+ c= 4a+ 2b+ c= 43.

That is, you have 4a- 2b+ c= 7, c= 13, and 4a+ 2b+ c= 43. It should be easy to solve those for a, b, and c and so write the quadratic function.

Of course, if $\displaystyle f(x)= ax^2+ bx+ c$, then $\displaystyle f'(x)= 2ax+ b$ and $\displaystyle f"(x)= 2a$. Evaluate those at x= 0 and x= 1.

Now, that does not use the values at x= 6, or 8 at all. You**could**, if you wished, write a $\displaystyle 4^{th}$ degree polynomial passing through those 5 points.

Again, the question, as you posed it, is impossible. There exist an infinite number of twice-differentiable functions passing through those 5 points. Unless there is more to this problem than you have told us, what I am suggesting is the simplest way to get**an**answer.