Which function has the greatest rate of change over the interval [0, 4]. y = x² – 2 y = 2 y = 2x + 1 They are all the same.

Which function has the greatest rate of change over the interval [0, 4].
y = x² – 2
y = 2
y = 2x + 1
They are all the same.
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  1. Answerh(x) has a greater average rate of change from x = 3π/2 to x =  2π======explanationi assume that when this question refers to "rate of change," it really means "average rate of change."the average rate of a change of a function f from x = a to x = b is given by    [tex]\dfrac{f(b) - f(a)}{b-a}[/tex]this expression applies for all functions, not just function f.  note that from the table, f(2π) = 0 and f(3π/2) = -2.the average rate of change of function f from  x = 3π/2 to x =  2π is    [tex]\begin{aligned} \dfrac{f(b) - f(a)}{b-a} & = \dfrac{f(2\pi) - f(3\pi/2)}{2\pi - \frac{3\pi}{2} } \\ & =\frac{0 - (-2)}{2\pi - \frac{3\pi}{2} } \\ & = \frac{2}{2\pi - \frac{3\pi}{2} }\\& \approx1.273 \end{aligned}[/tex]note that from the graph, g(2π) = 0 and g(3π/2) = -2.the average rate of change of function g from x = 3π/2 to x = 2π is    [tex]\begin{aligned} \dfrac{g(b) - g(a)}{b-a} & = \dfrac{g(2\pi) - g(3\pi/2)}{2\pi - \frac{3\pi}{2} } \\ & =\frac{0 - (-2)}{2\pi - \frac{3\pi}{2} } \\ & = \frac{2}{2\pi - \frac{3\pi}{2} }\\& \approx1.273\end{aligned}[/tex](the average rate of changes for functions f and g are the exact same.)note that from the equation, h(2π) = 6sin(2π) + 1 = 1  and h(3π/2) = 6sin(3π/2) + 1 = -5the average rate of change of function h from  x = 3π/2 to x =  2π is    [tex]\begin{aligned} \dfrac{h(b) - h(a)}{b-a} & = \dfrac{h(2\pi) - h(3\pi/2)}{2\pi - \frac{3\pi}{2} } \\ & =\frac{0 - (-5)}{2\pi - \frac{3\pi}{2} } \\ & = \frac{5}{2\pi - \frac{3\pi}{2} } \\ & \approx 3.183\end{aligned}[/tex]h(x) has a greater average rate of change from x = 3π/2 to x =  2π

  2. The average rate of change for the function f(x) can be calculated from the following equation
    [tex]\frac{f( x_{2})-f( x_{1} )}{ x_{2} - x_{1} }[/tex]

    By applying the last formula on the given equations 
    (1) the first function f
    from the table f(3π/2) = -2   and   f(2π) = 0
    ∴ The average rate of f = [tex]\frac{f(2 \pi)-f( \frac{3 \pi}{2} )}{2 \pi - \frac{3 \pi}{2} } = \frac{0-(-2)}{ \frac{\pi}{2} }= \frac{2}{ \frac{\pi}{2} } = \frac{4}{\pi}[/tex]

    (2) the second function g(x)
    from the graph g(3π/2) = -2   and   g(2π) = 0
    ∴ The average rate of g = [tex]\frac{g(2 \pi)-g( \frac{3 \pi}{2} )}{2 
\pi - \frac{3 \pi}{2} } = \frac{0-(-2)}{ \frac{\pi}{2} }= \frac{2}{ 
\frac{\pi}{2} } = \frac{4}{\pi}[/tex]

    (3) the third function h(x) = 6 sin x +1
    ∴ h(3π/2) = 6 sin (3π/2) + 1 = 6 *(-1) + 1 = -5
       h(2π) = 6 sin (2π) + 1 = 6 * 0 + 1 = 1
    ∴ The average rate of h = [tex]\frac{f(2 \pi)-f( \frac{3 \pi}{2} )}{2 
\pi - \frac{3 \pi}{2} } = \frac{1-(-5)}{ \frac{\pi}{2} }= \frac{6}{ 
\frac{\pi}{2} } = \frac{12}{\pi}[/tex]

    By comparing the results, The function which has the greatest rate of change is h(x)

    So, the correct answer is option C) h(x)

  3. Answer
    h(x) has a greater average rate of change from x = 3π/2 to x =  2π

    explanation

    i assume that when this question refers to "rate of change," it really means "average rate of change."

    the average rate of a change of a function f from x = a to x = b is given by

       

    this expression applies for all functions, not just function f. 

    note that from the table, f(2π) = 0 and f(3π/2) = -2.
    the average rate of change of function f from  x = 3π/2 to x =  2π is

       

    note that from the graph, g(2π) = 0 and g(3π/2) = -2.
    the average rate of change of function g from x = 3π/2 to x = 2π is

       

    (the average rate of changes for functions f and g are the exact same.)

    note that from the equation,
    h(2π) = 6sin(2π) + 1 = 1  and h(3π/2) = 6sin(3π/2) + 1 = -5
    the average rate of change of function h from  x = 3π/2 to x =  2π is

       

    h(x) has a greater average rate of change from x = 3π/2 to x =  2π

  4. Remember that the average rate of change of a function over an interval is the slope of the straight line connecting the end points of the interval. To find those slopes, we are going to use the slope formula: [tex]m= \frac{y_{2}-y_{1}}{x_2-x_1}[/tex]

    Rate of change of [tex]a[/tex]:
    From the graph we can infer that the end points are (0,1) and (2,4). So lets use our slope formula to find the rate of change of [tex]a[/tex]:
    [tex]m= \frac{y_{2}-y_{1}}{x_2-x_1}[/tex]
    [tex]m= \frac{4-1}{2-0}[/tex]
    [tex]m= \frac{3}{2}[/tex]
    [tex]m=1.5[/tex]
    The average rate of change of the function [tex]a[/tex] over the interval [0,2] is 1.5

    Rate of change of [tex]b[/tex]:
    Here the end points are (0,0) and (2,2)
    [tex]m= \frac{2-0}{2-0}[/tex]
    [tex]m= \frac{2}{2}[/tex]
    [tex]m=1[/tex]
    The average rate of change of the function [tex]b[/tex] over the interval [0,2] is 1

    Rate of change of [tex]c[/tex]:
    Here the end points are (0,-1) and (2,0)
    [tex]m= \frac{0-(-1)}{2-0}[/tex]
    [tex]m= \frac{1}{2}[/tex]
    [tex]m=0.5[/tex]
    The average rate of change of the function [tex]c[/tex] over the interval [0,2] is 0.5

    Rate of change of [tex]d[/tex]:
    Here the end points are (0,0.5) and (2,2.5)
    [tex]m= \frac{2.5-0.5}{2-0}[/tex]
    [tex]m= \frac{2}{2}[/tex]
    [tex]m=1[/tex]
    The average rate of change of the function [tex]d[/tex] over the interval [0,2] is 1

    We can conclude that the function that has the greatest rate of change over the interval [0, 2] is the function a.

  5. The function that has the greatest rate of change is:

                         g(x)

    Step-by-step explanation:

    We know that the rate of change from x=a to x=b is determined as:

    [tex]\dfrac{f(b)-f(a)}{b-a}[/tex]

    We are asked to find the rate of change of each of the functions form x=0 to x=pi over 2.

    f(x):

    We are given that:

    [tex]f(\dfrac{\pi}{2})=2[/tex]

    and,

    [tex]f(0)=0[/tex]

    Hence,

    [tex]rate\ of\ change=\dfrac{2-0}{\dfrac{\pi}{2}-0}\\\\\\rate\ of\ change=\dfrac{4}{\pi}[/tex]

    g(x):

    We have:

    [tex]g(0)=0[/tex]

    and

    [tex]g(\dfrac{\pi}{2})=4[/tex]

    Hence,

    [tex]rate\ of\ change=\dfrac{4-0}{\dfrac{\pi}{2}-0}\\\\\\rate\ of\ change=\dfrac{8}{\pi}[/tex]

    h(x):

    [tex]h(x)=\sin (x-\pi)+5[/tex]

    Now we have:

    [tex]h(0)=\sin (-\pi)+5\\\\h(0)=0+5\\\\h(0)=5[/tex]

    Also,

    [tex]h(\dfrac{\pi}{2})=\sin (\dfrac{\pi}{2}-\pi)+5\\\\\\h(\dfrac{\pi}{2})=\sin (\dfrac{-\pi}{2})+5\\\\h(\dfrac{\pi}{2})=-\sin (\dfrac{\pi}{2})+5\\\\\\h(\dfrac{\pi}{2})=-1+5\\\\h(\dfrac{\pi}{2})=4[/tex]

    Hence, the rate of change is calculated as:

    [tex]rate\ of\ change=\dfrac{4-5}{\dfrac{\pi}{2}-0}\\\\\\rate\ of\ change=\dfrac{-2}{\pi}[/tex]

                   Hence, the greatest rate of change is:

                                    g(x)

    Since,

    [tex]\dfrac{8}{\pi}\dfrac{4}{\pi}\dfrac{-2}{\pi}[/tex]

  6. Remember that the average rate of change of a function over an interval is the slope of the straight line connecting the end points of the interval. To find those slopes, we are going to use the slope formula: [tex]m= \frac{y_{2}-y_{1}}{x_2-x_1}[/tex]

    Rate of change of [tex]a[/tex]:
    From the graph we can infer that the end points are (0,1) and (2,4). So lets use our slope formula to find the rate of change of [tex]a[/tex]:
    [tex]m= \frac{y_{2}-y_{1}}{x_2-x_1}[/tex]
    [tex]m= \frac{4-1}{2-0}[/tex]
    [tex]m= \frac{3}{2}[/tex]
    [tex]m=1.5[/tex]
    The average rate of change of the function [tex]a[/tex] over the interval [0,2] is 1.5

    Rate of change of [tex]b[/tex]:
    Here the end points are (0,0) and (2,2)
    [tex]m= \frac{2-0}{2-0}[/tex]
    [tex]m= \frac{2}{2}[/tex]
    [tex]m=1[/tex]
    The average rate of change of the function [tex]b[/tex] over the interval [0,2] is 1

    Rate of change of [tex]c[/tex]:
    Here the end points are (0,-1) and (2,0)
    [tex]m= \frac{0-(-1)}{2-0}[/tex]
    [tex]m= \frac{1}{2}[/tex]
    [tex]m=0.5[/tex]
    The average rate of change of the function [tex]c[/tex] over the interval [0,2] is 0.5

    Rate of change of [tex]d[/tex]:
    Here the end points are (0,0.5) and (2,2.5)
    [tex]m= \frac{2.5-0.5}{2-0}[/tex]
    [tex]m= \frac{2}{2}[/tex]
    [tex]m=1[/tex]
    The average rate of change of the function [tex]d[/tex] over the interval [0,2] is 1

    We can conclude that the function that has the greatest rate of change over the interval [0, 2] is the function a.

  7. The function that has the greatest rate of change is:

                         g(x)

    Step-by-step explanation:

    We know that the rate of change from x=a to x=b is determined as:

    [tex]\dfrac{f(b)-f(a)}{b-a}[/tex]

    We are asked to find the rate of change of each of the functions form x=0 to x=pi over 2.

    f(x):

    We are given that:

    [tex]f(\dfrac{\pi}{2})=2[/tex]

    and,

    [tex]f(0)=0[/tex]

    Hence,

    [tex]rate\ of\ change=\dfrac{2-0}{\dfrac{\pi}{2}-0}\\\\\\rate\ of\ change=\dfrac{4}{\pi}[/tex]

    g(x):

    We have:

    [tex]g(0)=0[/tex]

    and

    [tex]g(\dfrac{\pi}{2})=4[/tex]

    Hence,

    [tex]rate\ of\ change=\dfrac{4-0}{\dfrac{\pi}{2}-0}\\\\\\rate\ of\ change=\dfrac{8}{\pi}[/tex]

    h(x):

    [tex]h(x)=\sin (x-\pi)+5[/tex]

    Now we have:

    [tex]h(0)=\sin (-\pi)+5\\\\h(0)=0+5\\\\h(0)=5[/tex]

    Also,

    [tex]h(\dfrac{\pi}{2})=\sin (\dfrac{\pi}{2}-\pi)+5\\\\\\h(\dfrac{\pi}{2})=\sin (\dfrac{-\pi}{2})+5\\\\h(\dfrac{\pi}{2})=-\sin (\dfrac{\pi}{2})+5\\\\\\h(\dfrac{\pi}{2})=-1+5\\\\h(\dfrac{\pi}{2})=4[/tex]

    Hence, the rate of change is calculated as:

    [tex]rate\ of\ change=\dfrac{4-5}{\dfrac{\pi}{2}-0}\\\\\\rate\ of\ change=\dfrac{-2}{\pi}[/tex]

                   Hence, the greatest rate of change is:

                                    g(x)

    Since,

    [tex]\dfrac{8}{\pi}\dfrac{4}{\pi}\dfrac{-2}{\pi}[/tex]

  8. F(x) sine curve with points at 0, 0 and pi over 2, 4 and pi, 0 and 3 pi over 2, negative 4 and 2 pi,0

    g(x) xy 00 pi over 22 π0 3 pi over 2−2 2π0

    h(x) = 2 sin x + 3 Which function has the greatest rate of change on the interval from x = 0 to x = pi over 2

    The change of f(x) from 0 to π/2 is 4
    The change of g(x) from 0 to π/2 is 2
    We can rule out g(x). 

    As for h(x):
    h(0) = 2 sin(0) + 3 = 3
    h(π/2) = 2(sin(π/2)) + 3 = 2 + 3 = 5
    Change of h(x) from 0 to π/2 is 2. 

    Greatest change between 0 and π/2 is found with f(x)

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