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3 years ago

Find the domain of the function:

Mathematics

2 answers:

makkiz [27]3 years ago

8 0

Answer: OPTION C.

Step-by-step explanation:

1. The domain of the function given in the problem are all those values for which the function that is in the denomiantor is different from zero, because the division by 0 is not allowed.

2. You can make the denominator equal to zero and solve it, as you can see below:

$x^{3}-2x^{2}+x=0\\x(x^{2}-2x+1)=0\\x(x-1)(x-1)=0\\x=0\\x=1$

3. Therefore, the domain is:

(-∞,0)U(0,1)U(1,∞)

ANEK [815]3 years ago

3 0

Answer:

Option C is correct.

Step-by-step explanation:

We have given the function.

f(x) = (x-1)²/(x³-2x²+x)

We have to find the domain of given function.

domain of function is all possible values of independent variable for which the function is difined.

So for this the denominator must not be zero.

x³-2x²+x ≠ 0

x(x²-2x+1)≠ 0

x(x-1)(x-1) ≠ 0

x ≠0 and x≠1

So, the domain cannot consist of x=0 and x=1.

Therefore, the domain is :

(-∞ , 0) ∪ (0,1 ) ∪ (1,+∞).

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Find the first partial derivatives of the function f(x,y,z)=4xsin(y−z)

Amanda [17]

Answer:

$f_x(x,y,z)=4\sin (y-z)$

$f_x(x,y,z)=4x\cos (y-z)$

$f_z(x,y,z)=-4x\cos (y-z)$

Step-by-step explanation:

The given function is

$f(x,y,z)=4x\sin (y-z)$

We need to find first partial derivatives of the function.

Differentiate partially w.r.t. x and y, z are constants.

$f_x(x,y,z)=4(1)\sin (y-z)$

$f_x(x,y,z)=4\sin (y-z)$

Differentiate partially w.r.t. y and x, z are constants.

$f_y(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial y}(y-z)$

$f_y(x,y,z)=4x\cos (y-z)$

Differentiate partially w.r.t. z and x, y are constants.

$f_z(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial z}(y-z)$

$f_z(x,y,z)=4x\cos (y-z)(-1)$

$f_z(x,y,z)=-4x\cos (y-z)$

Therefore, the first partial derivatives of the function are $f_x(x,y,z)=4\sin (y-z), f_x(x,y,z)=4x\cos (y-z)\text{ and }f_z(x,y,z)=-4x\cos (y-z)$ .