Let f(x)=sin(x)+cos(2x). Find a quadratic polynomial p(x) so that p(0)=f(0), p'(0)=f'(0) and p''(0)=f''(0).
Hello Velvet Love
First, find the derivatives of f(x)
It is f'(x) = cos(x)-2sin(2x) and
f''(x) = -sin(x) -4cos(2x)
p(x) should be quadratic, so
p(x) = ax^2+bx+c
In this excercise we have to find a,b,c and want to work with derivatives, so let's find p'(x) and p''(x) first
p'(x) = 2ax+b
p''(x) = 2a
( It should be p(0)=f(0), p'(0)=f'(0) and p''(0)=f''(0) )
I suggest we start with the second derivative, because p''(x) has only one unknown in it - the 'a'. p(x) contains a,b and c.
So
p''(0) = f''(0)
2a = -sin(0) -4cos(2*0)
2a = 0 -4cos(0)
2a = -4*1
=> a = -2.
So p'(x) = 2ax+b = 2*(-2)a+b
Now consider
p'(0) = f'(0) and solve for b.
then p(0) = f(0) and solve for c.
Done
Do you have some question on that?
Yours
Rapha