The value of [tex]f_{x}[/tex](3,2)≈12.2, [tex]f_{x}[/tex](3,2.2)≈16.8, [tex]f_{x}[/tex](3,2)≈23.25
This problem aims to find the values of a function having alternate independent variables.
A-
[tex]f_{x}(3,2) f_{x} (x,y)=lim_{h-0} \frac{(3+0.5,2)-f(3,2)}{0.5}[/tex] considering h =±0.5
Solving for h=0.5
⇒[tex]\frac{f(3.5,2)-f(3,2)}{0.5}[/tex]
Using the table to plug in the values of the function: 22.4-17.5/0.5=9.8
Now, solving for h=-0.5
⇒[tex]\frac{f(3.5,2)-f(3,2)}{-0.5}[/tex]
Using the table to plug in the values of the function: 22.4-17.5/(-0.5)=14.6
Taking the average of both ±0.5 answers for the final answer of f(3,2)
[tex]f_{x}(3,2)=\frac{9.8+14.6}{2} \\or, f_{x}(3,2)=12.2[/tex]
B-
[tex]f_{x}(3,2.2)=lim_{h-0} \frac{(3+0.5,2.2)-f(3,2.2)}{0.5}[/tex] considering h =±0.5
Solving for h=0.5
⇒[tex]\frac{f(3.5,2.2)-f(3,2.2)}{-0.5}[/tex]
Using the table to plug in the values of the function: 26.1-15.9/0.5=20.4
Now, solving for h=-0.5
⇒[tex]\frac{f(2.5,2.2)-f(3,2.2)}{-0.5}[/tex]
Using the table to plug in the values of the function: 9.3-15.9/(-0.5)=13.2
Taking the average of both ±0.5 answers for the final answer of f(3,2.2)
[tex]f_{x}(3,2.2)=\frac{20.4+13.2}{2} \\or, f_{x}(3,2.2)=16.8[/tex]
C-
[tex]f_{xy} (3,2)=lim_{h-0} \frac{f_{x}(3,2+h)-f_{x} (3,2) }{h} \\[/tex] considering h =±0.2
Solving for h=0.2
⇒[tex]\frac{f_{x} (3,2.2)-f_{x} (3,2)}{0.2}[/tex]
Plugging in the answers from A and B: 16.8-12.2/0.2= 23
Now solving for h= -0.2
⇒[tex]\frac{f_{x} (3,1.8)-f_{x} (3,2)}{-0.2}[/tex]
Plugging in the values: 7.5-12.2/-0.2= 23.5
Taking an average of h=±0.2 answers to find the final answer:
[tex]f_{x}(3,2)=\frac{23+23.5}{2} \\or, f_{x}(3,2)=23.25[/tex]
To know more about limits visit: brainly.com/question/8533149
#SPJ4
