You confused me with having 3 coordinates in f(x,y).
I would use a=4 and b=0.
you want
Last one, guys...
"Given the function f(x,y)=x^2cos(xy)-5xy, estimate the value of f(3,97,0.025)."
My work:
fx(x,y) = 2x(cos(xy))-y(x^2sin(xy)+5)
fy(x,y) = -x(x^2sin(xy)+5)
So, f(x,y) + fx(x,y)(x-a) + fy(x,y)(y-b)
f(3.97,.025) + fx(3.97,.025)(3.97-3.95) + fy(x,y)(.025-.020)
And, after plugging into this equation, my final answer was 15.32.
Did I mess up anywhere? I'm pretty confident about the answer, since (3.97,.025) plugged into the original equation is 15.26, but I just want to make sure I'm not missing anything.