Answer: The approximate f(1.1,-0.1) is 4.92.
Step-by-step explanation:
Since we have given that
[tex]f(x,y)=5xe^{xy}[/tex]
at (1,0),
we get [tex]f(1,0)=5[/tex]
Partial derivative would be
[tex]f_x=5e^{xy}+5xye^{xy}=5e^{xy}(1+xy)[/tex]
So,
[tex]f_x(1,0)=5[/tex]
Similarly,
[tex]f_y=5x^2ye^{xy}[/tex]
So, [tex]f_y(1,0)=5[/tex]
Now,
[tex]L(x,y)=f(1,0)+f_x(1,0)(x-1)+f_y(1,0)(y-0)\\\\L(x,y)=5+5(x-1)+5y\\\\L(x,y)=5+5x-5+5y\\\\L(x,y)=5x+5y[/tex]
L(1.1,-0.1)=[tex]5(1.1)-5(0.1)=5[/tex]
and
f(1.1,-0.1) = [tex]5(1.1)e^{1.1\times -0.1}=4.92[/tex]
Hence, the approximate f(1.1,-0.1) is 4.92.
