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zhenek [66]
3 years ago
12

Which points lie on the graph of f(x) = log9x? Check all that apply.

Mathematics
2 answers:
QveST [7]3 years ago
7 0

Answer: C, E, F

C) (1/9),-1

E) (9,1)

F) (81,2)

valkas [14]3 years ago
5 0

Answer:

Let

y=f(x)

so

y=log9x

we know that

applying property of logarithms

y= log9x is equal to

------> equation 1

so

case 1) (-1/81, 2)

x=-1/81

y=2

substitute the value of y in the equation 1 to obtain the value of x

81 is not equal to -1/81-------> the point does not belong to the graph

case 2) (0, 1)

x=0

y=1

substitute the value of y in the equation 1 to obtain the value of x

9 is not equal to 0-------> the point does not belong to the graph

case 3) (1/9, -1)

x=1/9

y=-1

substitute the value of y in the equation 1 to obtain the value of x

1/9 is equal to 1/9-------> the point belongs to the graph

case 4) (3, 243)

x=3

y=243

substitute the value of y in the equation 1 to obtain the value of x

9^{243} is not equal to 3-------> the point does not belong to the graph

case 5) (9, 1)

x=9

y=1

substitute the value of y in the equation 1 to obtain the value of x

9 is equal to 9-------> the point belongs to the graph

case 6) (81, 2)

x=81

y=2

substitute the value of y in the equation 1 to obtain the value of x

81 is equal to 81-------> the point belongs to the graph

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Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

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2. Functions

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That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2)\Re\geq0\rightarrow\Re , f(x)=x+1\\

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3)f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)

The cosine function is not injective, bijective nor surjective.

4)f:\Re\rightarrow\Re \:f(x)=ex

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5)  Quite unclear format

6) \:f:\Re\rightarrow(0,\infty), f(x) =ex

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

7) f:\Re\geq 0 \rightarrow  \Re\geq0, f(x) =x^{4}

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8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}

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10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.

Surjective.

11.f:R\geq 0[-1,1], f(x) = 0\\

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7 0
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4 years ago
Find dy/dx when r=2 2cos(theta) , then find slope of tengent line at point(4, 2pi)
Korolek [52]

The slope of tangent line at point(4, 2pi) is undefined.

For given question,

We have been given a polar equation r = 2 + 2cos(θ)

We need find dy/dx as well as the slope of tangent line at point(4, 2π).

We know that, for polar equation we use,

x = r cos(θ)   and  y = r sin(θ)

plug the given value of r into these equations we get:

⇒ x = r cos(θ)

⇒ x = (2 + 2cos(θ) ) ×  cos(θ)

⇒ x = 2(cos(θ) + cos²(θ))

⇒ x = 2cos(θ) + 2cos²(θ)

Similarly,

⇒ y = r sin(θ)

⇒ y = (2 + 2cos(θ) ) ×  sin(θ)

⇒ y =  2(sin(θ) + sin(θ)cos(θ))

⇒ y =  2sin(θ) + 2sin(θ)cos(θ)

Now we find derivative of x and y with respect to theta.

\Rightarrow \frac{dx}{d\theta} =-2sin(\theta)+2(-2cos(\theta)sin(\theta))\\\\\Rightarrow  \frac{dx}{d\theta} =-2sin(\theta)-2sin(2\theta)          .............(1)

Similarly,

\Rightarrow \frac{dy}{d\theta}=2cos(\theta)+2(cos^2(\theta)-sin^2(\theta))\\\\\Rightarrow \frac{dy}{d\theta}=2cos(\theta)+2(cos(2\theta))          ..............(2)

Now we find dy/dx

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From (1) and (2),

\Rightarrow \frac{dy}{dx} =\frac{2cos(\theta)+2cos(2\theta)}{-2sin(\theta)-2sin(2\theta)} \\\\\Rightarrow \frac{dy}{dx} =\frac{2(cos(\theta)+cos(2\theta))}{-2(sin(\theta)+sin(2\theta))}\\\\\Rightarrow \frac{dy}{dx} =-\frac{cos(\theta)+cos(2\theta)}{sin(\theta)+sin(2\theta)}

We know that The slope of tangent line is given by dy/dx.

So, the slope is: m =-\frac{cos(\theta)+cos(2\theta)}{sin(\theta)+sin(2\theta)}

Now we need to find the slope of tangent line at point(4, 2pi)

Substitute θ = 2π in above slope formula.

\Rightarrow m =-\frac{cos(2\pi)+cos(2\times 2\pi)}{sin(2\pi)+sin(2\times 2\pi)}\\\\\Rightarrow m=-\frac{1+cos(4\pi)}{0+sin(4\pi)}\\\\\Rightarrow m=-\frac{1+cos(4\pi)}{sin(4\pi)}

⇒ m = ∞

The slope of tangent line at point(4, 2pi) is not defined.

This means, the tangent line must be parallel to Y-axis.

Therefore, the slope of tangent line at point(4, 2pi) is undefined.

Learn more about the slope here:

brainly.com/question/10785137

#SPJ4

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