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Zielflug [23.3K]
3 years ago
5

Type the correct answer in the box. Use numerals instead of words.

Mathematics
1 answer:
Brums [2.3K]3 years ago
4 0

Answer:

824 feet

Step-by-step explanation:

Got it right on Plato

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9) Which of the following shows the Associative Property of Addition
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3 0
3 years ago
Let f be the function defined by f(x) = 3x^5-5x^3+2
sveticcg [70]
(a) When f is increasing the derivative of f is positive.

    f'(x) = 15x^4 - 15x^2 > 0
            15x^2(x^2 - 1)> 0
               x^2 - 1    > 0 (The inequality doesn't flip sign since x^2 is positive)
               x^2 > 1
    Then f is increasing when x < -1 and x > 1.

(b) The f is concave upward when f''(x) > 0.
    f''(x) = 60x^3 - 30x > 0
           30x(2x^2 - 1) > 0
             x(2x^2 - 1) > 0
            x(x^2 - 1/2) > 0
x(x - 1/sqrt(2))(x + 1/sqrt(2)) > 0

    There are four regions here. We will check if f''(x) > 0.
    x < -1/sqrt(2):     f''(-1) = -30 < 0
    -1/sqrt(2) < x < 0: f''(-0.5) = 7.5 > 0
    0 < x < 1/sqrt(2):  f''(0.5) = -7.5 < 0
    x > 1/sqrt(2):      f''(1) = 30 > 0

    Thus, f''(x) > 0 at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
    Therefore, f is concave upward at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).

(c) The horizontal tangents of f are at the points where f'(x) = 0
    15x^2(x^2 - 1) = 0
    x^2 = 1
    x = -1 or x = 1
    f(-1) = 3(-1)^5 - 5(-1)^3 + 2 = 4
    f(1) = 3(1)^5 - 5(1)^3 + 2 = 0

    Therefore, the tangent lines are y = 4 and y = 0.
8 0
4 years ago
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