Question

NO LINKS!!!

Please help me with #53.

Answer 1

53

a. Graph of equation attached

b.
Domain:

`-\infty \: < x < \infty \:`
Interval notation:
`\left(-\infty \:,\:\infty \:\right)`

Range:

`f(x) > 1\\\\`

Interval notation:
`\left(1,\:\infty \:\right)`

c. See explanation below and on attached graph

d. See explanation below

Explanation:

53

The function to be used for answering questions is

`f(x) = 8\left((1)/(4)\right)^(x+4)+1`

a. Graph

Graph is plotted using online graphing tool. 2 points identified are (-4, 9) and (-3, 3)

b. Domain and Range

First let's understand what domain and range are

Definition of domain

The domain of a function, y = f(x), is the set of input(x) values for which the function is real and defined

The function has no undefined points nor domain constraints. You can see that the graph goes to infinity for negative and positive x values Therefore, the domain is :

`-\infty \: < x < \infty \:`

which can be expressed in interval notation as:

`\left(-\infty \:,\:\infty \:\right)`

Definition of range

The range of a function y = f(x) is the set of values for y over the domain for which the function is defined

As
`x \rightarrow \infty`,
`y \rightarrow 1`

y never quite reaches 0 but its value approaches 1 and therefore the range is

`f(x) > 1\\\\`

In interval notation:

`\left(1,\:\infty \:\right)`

c. End Behavior

The end behavior of a function f(x) refers to how the function behaves when the variable x increases or decreases without bound. In other words, the end behavior describes the ultimate trend in the graph of f(x) as
`x \rightarrow \pm \infty`

In this particular function we see that as
`x\to \:+\infty \:,\:f\left(x\right)\to \:1\\`

and as
`x\to \:-\infty \:,\:f\left(x\right)\to \:\infty\\`

d. Point (-3, 3)
By pure chance I chose this point as one of the points asked for in part a because I was looking for integer values I don't think it matters that there is repetition

(-3, 3) is plotted on the graph as point A

3. Evaluate f(-3)
We have

`f(x) = 8\left((1)/(4)\right)^(x+4)+1`

To find f(-3) substitute -3 for x in the above function:

`f(-3) = 8\left((1)/(4)\right)^(-3+4)+1\\\\= 8\left((1)/(4)\right)^1} + 1\\\\\\= 8\left((1)/(4)\right) + 1\\\\= 2 + 1 \\\\= 3`

So f(-3) = 3

Answer 2

a) See attachment 1.

b) Domain: (-∞, ∞)

Range: (1, ∞)

c) As x → -∞, f(x) → ∞

As x → ∞, f(x) → 1⁺

d) See attachment 2.

e) See below.

Explanation:

Given exponential function:

`f(x)=8\left((1)/(4)\right)^(x+4)+1`

Part a

To find the y-intercept, input x = 0 and solve for y:

`\implies f(0)=8\left((1)/(4)\right)^(0+4)+1`

`\implies f(0)=8\left((1)/(4)\right)^(4)+1`

`\implies f(0)=8\left((1)/(256)\right)+1`

`\implies f(0)=(1)/(32)+1`

`\implies f(0)=(33)/(32)`

`\implies f(0)=1.03125`

Therefore, the y-intercept is (0, 1.03125).

Substitute x = -2 into the function to find a second point on the curve:

`\implies f(-2)=8\left((1)/(4)\right)^(-2+4)+1`

`\implies f(-2)=8\left((1)/(4)\right)^(2)+1`

`\implies f(-2)=8\left((1)/(16)\right)+1`

`\implies f(-2)=(1)/(2)+1`

`\implies f(-2)=(3)/(2)`

`\implies f(-2)=1.5`

Therefore, a second point on the curve is (-2, 1.5).

Part b

The domain of a function is the set of all possible input values (x-values).

The domain of the given function is unrestricted:

• Interval notation: (-∞, ∞)

The range of a function is the set of all possible output values (y-values).

As the exponential part of the function is always positive, the range of the given function is restricted:

• Interval notation: (1, ∞)

Part c

As x approaches negative infinity, the function approaches positive infinity:

• As x → -∞, f(x) → ∞

As x approaches positive infinity, the function approaches f(x) = 1.

• As x → ∞, f(x) → 1⁺

Part d

See attachment 2.

Part e

Substitute x = -3 into the function:

`\implies f(-3)=8\left((1)/(4)\right)^(-3+4)+1`

`\implies f(-3)=8\left((1)/(4)\right)^(1)+1`

`\implies f(-3)=8\left((1)/(4)\right)+1`

`\implies f(-3)=2+1`

`\implies f(-3)=3`