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iris [78.8K]
3 years ago
6

Maksim was asked on a test "What is the domain of the function: f(x) = √(x - 2)? Explain." Maksim responded with "The function f

(x) = √(x - 2) has a domain of all real numbers because any number can be put in for x." Explain if Maksim's answer and explanation were correct. Make sure to use at least two pieces of evidence to support your claim. If he is wrong, state the actual domain.
Mathematics
1 answer:
swat323 years ago
4 0

Answer:

Maksim's answer and explanation are not correct

The actual domain is 2 < x < Infinity

Step-by-step explanation:

The domain of a function is the set that is made up of the possible inputs of the function

The given function is f(x) = √(x - 2)

Maksim's response for the domain of f(x) is x ∈ R ('x' is a member of the set of all real numbers

The function is not defined when x < 2, at which the expression, x - 2, will become negative, and for which the square root of the negative number is imaginary

Therefore, the domain for which the function is defined is all real numbers larger than 2, which can be presented as follows;

The domain of f(x) = 2 < x < ∞

Therefore Maksim is wrong as the actual domain is limited to 2 < x < ∞.

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Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t
Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

\frac{dy}{dx}=0 (The slope is zero)

--To obtain vertical tangent lines, the condition is:

\frac{dy}{dx}=\frac{1}{0} (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}=2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0

Solving for dy/dx:

\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}

Applying the first conditon (slope is zero)

\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0

Solving for y (Adding 2x+4, dividing by 2)

y=x+2 (I)

Replacing (I) in the given equation:

x^{2}-2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0\\x^{2}-2x^{2}-4x+x^{2} +4x+4+4x-8x-16+20=0\\-4x+8=0\\x=2

Replacing it in (I)

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Therefore, the parabola has a horizontal tangent line at the point (2,4)

Applying the second condition (slope is undefined where denominator is zero)

2y-2x-8=0

Adding 2x+8 both sides and dividing by 2:

y=x+4(II)

Replacing (II) in the given equation:

x^{2}-2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0\\x^{2}-2x^{2}-8x+x^{2}+8x+16+4x-8x-32+20=0\\-4x+4=0\\x=1

Replacing it in (II)

y=1+4

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Answer:

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So time spent on puppy videos =  9 X 52

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