I really need some help I am lost and very bad at graphs. Thank you in advance

Question

 

f(x)

g(x) x y 0 0 2 π 0 −2 2π 0

h(x) = 2 sin x + 3

Which function has the greatest rate of change on the interval from x = 0 to x = ?

 f(x)

 g(x)

 h(x)

 All three functions have the same rate of change.

10) Use the graph below to answer the question that follows:

 Amplitude: 4; period: π; midline: y = 2

 Amplitude: 2; period: 2π; midline: y = 1

 Amplitude: 4; period: 2π; midline: y = 1

 Amplitude: 2; period: π; midline: y = 2

13) Which of the following graphs best represents f(x) = 3 cos(4x) + 1?

 

 

 

 

23) Use the graph below to answer the question that follows:

 

 

 

 

25) Compare each of the functions shown below:

f(x) = 3 cos(2x − π) − 2

g(x)

h(x) x y −2 −4 −1 1 0 4 1 5 2 4 3 1 4 −4

Which function has the largest maximum?

 f(x)

 g(x)

 h(x)

 All three functions have the same maximum value.

Answer 1

(7).

First find the average rate for each function at x = 0 and x = π/2.

Average rate for Function f(x):

The curve f(x) with the points (0, 0), (π/2, 4), (π, 0), (3π/2, -4) and (2π, 0).

Let the points are (x₁, y₁) = (0, 0) and (x₂, y₂) = (π/2, 4).

Average rate = dy/dx =( y₂ - y₁)/(x₂ - x₁) = (4 - 0) / (π/2 - 0) = 4/(π/2) = 8/π = 8 / (22/7) = 2.54.

 

Average rate for Function g(x):

The curve g(x) with the points (0, 0), (π/2, 2), (π, 0), (3π/2, -2) and (2π, 0).

Let the points are (x₁, y₁) = (0, 0) and (x₂, y₂) = (π/2, 2).

Average rate = dy/dx =( y₂ - y₁)/(x₂ - x₁) = (2 - 0) / (π/2 - 0) = 2/(π/2) = 4/π = 4 / (22/7) = 1.27.

 

Average rate for Function h(x):

h(x) = 2 sin(x) + 3

h(0) = 2 sin(0) + 3 = 3

h(π/2) = 2 sin(π/2) + 3 = 5

Average rate = dy/dx =( y2 - y1)/(x2 - x1) = (5 - 3) / (π/2 - 0) = 2/(π/2) = 4/π = 4 / (22/7) = 1.27.

 

Arrange the average rates least to greatest

1.27 = 1.27 and 2.54

g(x) = h(x), f(x)

The greatest rate change function is f(x).

Answer 2

10)

Given points are

(0,-1),(pi/3,0),(pi,3),(5pi/3,0),(2pi,-1)

Calculate the amplitude

A =  (maximum-minimum)/2

   =(3-(-1))/2

  =2

Period of cosine function is 2pi

So middle line of the function is y=1

Therefore, the option 2 is correct.

 

 

Answer 3

13)

The function is  f(x) = 3 cos (4x) + 1?

In four options

x values are 0,pi/4,pi/2

So calculate the y values.

At x=0

f(x) = 3 cos (4*0) + 1=3+1=4

At x=pi/4

f(x) = 3 cos (4*(pi/4)) + 1=-3+1=-2

At x=pi/2

f(x) = 3 cos (4*(pi/2)) + 1=3+1=4

Therefore, the first option is correct.

Answer 4

(25).

f(x) = 3 cos(2x-π) - 2

Any sine or cosine function that takes the form y = a*cos(bx + c) + d , since -1 ≤ cos(bx + c) ≤ 1:

-1 ≤ cos(2x-π) ≤ 1

- 3 ≤ 3 cos(2x-π) ≤ 3

- 3 - 2 ≤ 3 cos(2x-π) - 2 ≤ 3 - 2

- 5 ≤ 3 cos(2x-π) - 2 ≤ 1

Maximiu y value = 1

 

The graph of a downward facing parabola with vertex at (2, 3). So, the the function g(x) is maximum at (2, 3).

 

From the points of h(x), the maximum y-value at x = 1 is 5. So the function h(x) is maximum at (1, 5).

 

The function h(x) has the greatest maximum y value i.e = 5. So, the function h(x) is has the largest maximum.

 

Answer 5

(23).

First find the average rate for each function at x = π and x = 3π/2.

Average rate for the curve :

The curve with the points (0, 7), (π/2, -1), (π, 7) and (3π/2, -1).

Let the points are (x₁, y₁) = (π, 7) and (x₂, y₂) = (3π/2, -1).

Average rate = dy/dx =( y₂ - y₁)/(x₂ - x₁) = (-1 - 7) / (3π/2 - π) = - 8/(π/2) = - 16/π.

The second option is correct answer.