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3 years ago

I'm having a bit of trouble finding out how to find the Domain of this function. Help please!

Mathematics

1 answer:

BabaBlast [244]3 years ago

6 0

In general, the domain is the set of all x-values for the graph.

The issue here is that this isn't the graph of a function. A function has at most one y-value for each x-value and this graph has an infinite number of y-values for the single x-value of 1.

So, either your teacher is wanting you to say the domain is {1}, because that's the only x-value used by the graph, or they're wanting you to say this is a trick question, because this isn't the graph of a function.

The range is the set of all y-values, which is -9<y<9, but again, do they intend this to be a trick question?

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Find dy/dx for y= x^3 ln (cot x)

ICE Princess25 [194]

<h3>Answer</h3>

$\dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)$

<h3>Explanation</h3>

By the product rule $(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$ , we have

$\begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}$

By the chain rule:

$\begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}$

By the power rule:

$(x^3)' = 3x^2$

thus

$\begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}$

Nothing to do to simplify any further, other than factoring out $x^2$ .

4 0

4 years ago

Rewrite the fraction 5 over 4 as an equivalent fraction with a denominator equal to 16

irakobra [83]

5/4=
20/16
Answer: 20/16

4 0

3 years ago

Read 2 more answers

A plane is 24,000 feet above sea

IrinaVladis [17]

Answer:

A. -500

Step-by-step explanation:

Find the slope using rise over run (change in y / change in x)

Find the change in the y values (height of the plane):

24,000 - 18,000

= 6,000

Find the change in the x values (miles from the airport)

(60 - 72)

= -12

Divide 6000 by -12:

6000/-12

= -500

So, the slope of the plane's path is A. -500

5 0

3 years ago

Read 2 more answers

Point A is located at (3,-6) and point A is located at (1-2). What is the scale factor?

Anuta_ua [19.1K]

Answer:

The scale factor is 3.

Step-by-step explanation:

Let the origin O(0,0) is the reference point and the coordinates of the point A are given by (3,-6).

Therefore, the distance OA is given by $\sqrt{(3 - 0)^{2} + (- 6 - 0)^{2}} = 3\sqrt{5}$ units.

Again, the coordinates of point A' are (1,-2).

Therefore, the distance OA' is given by $\sqrt{(1 - 0)^{2} + (- 2 - 0)^{2}} = \sqrt{5}$ units.

Hence, the scale factor is $\frac{OA}{OA'} = \frac{3\sqrt{5} }{\sqrt{5}} = 3$ . (Answer)

4 0

3 years ago

Mandy shaded the fraction strip below to represent a fraction Shade the fraction strip below so that it represents a fraction th

Aleksandr-060686 [28]

Your question is missing the figure, so the figure for your question is attached below:

Answer:

shade 2 strips out of 4 to get fraction strip equivalent to Mandy's fraction strip

Step-by-step explanation:

As Mandy shaded the 3 trips out of the total six strips. It shows the fraction of $\frac{3}{6}$

and $\frac{3}{6}=\frac{1}{2}$

To shade the given fraction strip so that it represents a fraction that is equivalent to Mandy's fraction strip, we should shade 2 stripes out of 4 that is equivalent to $\frac{1}{2}$

i.e. $\frac{2}{4}=\frac{1}{2}$

My Fraction Strip is equivalent to Mandy's Fraction Strip because both are equal to $\frac{1}{2}$

5 0

4 years ago