a) True
The given function
f:R3→R is defined by f(x,y,z)=∣x∣+∣y∣+∣z∣.
Consider,the following maps.
f1:R3→R defined by f1(x,y,z)=∣x∣ ,
f2:R3→R defined by f2(x,y,z)=∣y∣ ,
f3:R3→R defined by f3(x,y,z)=∣z∣ .
Clearly,f1,f2,f3 are differentiable at (3,2,−1) ,since the absolute value functions are differentiable everywhere except (0,0,0).
Again, we known that sum of differentiable functions is again differentiable.
Therefore,f:R3→R defined by
f(x,y,z)=f1(x,y,z)+f2(x,y,z)+f3(x,y,z)=∣x∣+∣y∣+∣z∣
is differentiable at (3,2,−1) .
b) False.
A homogeneous real valued function of two variable x and y is a real valued function that satisfies the condition
f(rx,ry)=rkf(x,y) for some constant k and all real number r.
Now, the given real valued function of two variable is
f(x,y)=max{xy,x} .
Putting,x=rx and y=ry ,we get
f(rx,ry)=max(rxry,rx)=max(xy,rx)=rkf(x,y).
for some k and all real number r.
Hence,f is not a homogeneous function.
c) False.
Given function are
f:R2:→ R defined by f(x,y)=2xy.
g:R2:→R defined by g(x,y)=x2+y2 .
Now ,
gf(x,y)=x2+y22xybut at (0,0),
gf(0,0)=00
which is not defined.
Hence , R2 is not a domain of gf .
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