We know that: $$f (0) = 3$$, $$f'(0) = 1$$, $$f(1) = 2$$ and $$f'(1)=-2$$. Calculate the polynomial of Hermite that interpolates these points.
See development and solution
Development:
In this case we have $$n+1=2$$ points, therefore the degree of the polynomial of Hermite will be $$2n+1 = 3$$. We proceed as we explained, we write in a table the points repeating those in which we know the derivative:
| $$0$$ | $$3$$ | |||
| $${f'}_0=1$$ | ||||
| $$0$$ | $$3$$ | $$\dfrac{-1-1}{1-0}=-2$$ | ||
| $$\dfrac{2-3}{1}=-1$$ | $$\dfrac{-1+2}{1-0}=1$$ | |||
| $$1$$ | $$2$$ | $$\dfrac{-2+1}{1-0}=-1$$ | ||
| $${f'}_1=-2$$ | ||||
| $$1$$ | $$2$$ |
Then, the polynomial is written in the same way, taking the first element of every column (starting from the second).
$$$\begin{array}{rl} P_3(x)=& 3+1(x-0)-2(x-0)^2+1(x-0)^2(x-1)\\ =& 3+x-2x^2+x^3-x^2= x^3-3x^2+x+3 \end{array}$$$
Solution:
$$P_3(x)= x^3-3x^2+x+3 $$