Use the difference quotient of A(r)= πr2 to calculate the average rate of change during certain intervals.The difference quotient of A(r) is:StartFraction

Use the difference quotient of A(r)= πr2 to calculate the average rate of change during certain intervals. The difference quotient of A(r) is:

StartFraction pi (r + h) squared minus pi r squared Over h EndFraction = StartFraction pi (r squared + 2 r h + h squared) minus pi r squared Over h EndFraction = StartFraction pi r squared + 2 pi r h + pi h squared minus pi r squared Over h EndFraction = 2 pi r + pi h

The average rate of change from the interval 2 to 3 inches is
π square inches per inch.

The average rate of change from the interval 2 to 2.5 is
π square inches per inch.

Related Posts

This Post Has 7 Comments

  1. [tex]f(x)=x^2\\\\\dfrac{f(x+h)-f(x)}{h}=\dfrac{(x+h)^2-x^2}{h}\\\\=\dfrac{x^2+2xh+h^2-x^2}{h}=\dfrac{2xh+h^2}{h}\\\\=\dfrac{h(2x+h)}{h}=\boxed{2x+h}[/tex]

    Used:
    (a + b)² = a² + 2ab + b²

  2. Substitute –3 for x and 5 for h in the difference quotient.

    Step-by-step explanation:

    The difference quotient for the function f(x) is  21x² + 21xh + 7h² + 2. Now, we know that the difference quotient equal f(x + h) - f(x) = 21x² + 21xh + 7h² + 2. The change in x is h = x₂ - x₁. So, if x changes from x = -3 to x = 2,  where x₁ = -3 and x₂ = 2, h = x₂ - x₁ = 2 - (-3) = 2 + 3 = 5.

    So to find the average rate of change of f(x) from x = -3 to x = 2, we substitute x = -3 and h = 5 into the difference equation  f(x + h) - f(x) = 21x² + 21xh + 7h² + 2. Since, x starts at x = -3 and increases by 5 units to x = 2.

  3. 3. (5x + 1) - (-10x + 6)

    simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

    - (-10x + 6) = + 10x - 6

    simplify. combine like terms:

    5x + 1 + 10x - 6

    5x + 10x + 1 - 6

    15x - 5

    ∴ it is your answer

    ~

  4. Ok, we have:

    d(t) = 3*t^2 + 5*t - 2.

    The first interval is:

    (2, 3)

    and remember that, for an interval (a,b), the difference quotient is:

    D = (f(b) - f(a))/(b -a)

    Then in this first interval we have:

    [tex]d = \frac{d(3) - d(2)}{3-2} = \frac{3*3^2 + 5*3 - 2 - (3*2^2 + 5*2 - 2)}{1} = 20[/tex]

    So the average rate of change in (2,3) is 20.

    now, in (2, 2.5) we have:

    [tex]d = \frac{d(2.5) - d(2)}{3-2} = \frac{3*2.5^2 + 5*2.5 - 2 - (3*2^2 + 5*2 - 2)}{1} = 9.25[/tex]

    So here the rate of change is 9.25

    And in the interval (2, 2.1) we have:

    [tex]d = \frac{d(2.1) - d(2)}{3-2} = \frac{3*2.1^2 + 5*2.1 - 2 - (3*2^2 + 5*2 - 2)}{1} = 1.73[/tex]

    So in this interval the rate of change is 1.73

Leave a Reply

Your email address will not be published. Required fields are marked *