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Find the functional values f(-2), f(0), and f(1) for the function. ƒ(x) = x² - 1
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Find the functional values f(-2), f(0), and f(1) for the function. ƒ(x) = X²- 1
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To find the functional values f(-2), f(0), and f(1) for the function ƒ(x) = x² - 1, you simply
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Find the functional values f(-2), f(0), and f(1) for the function. ƒ(x) = x² - 1
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f(-2)= (-2)^2 - 1 = 3 Can you do the rest?
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Find the functional values f(-5), f(1), and f(3) for the compound function.
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f(-5) = |-5-5| = 10 , use the first f(1) = |1-5| = 4 , use the first f(3) = 1/4 , use the second
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Find the functional values f(-5), f(1), and f(3) for the compound function.
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|-10| = 10 |-4| = 4 1/4
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f = (3/4)x^4-x^3-3x^2+6x f' = 3x^3-3x^2-6x+6 f" = 9x^2-6x-6 extrema where f'=0 and f"≠0 f
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1.Consider the function f(x)= (3/4)x4 - x3 - 3x2 +6x
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#1 Extrema occur where the derivative is zero f(x)= (3/4)x^4 - x^3 - 3x^2 +6x f'(x) = 3x^3 - 3x^2 -
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Consider the function f(x) = 1/4x^4 - 5/3x^3 - 3/2x^2 + 15x - 3.
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When you say, "I don't understand any of this", I am troubled. This question is as fundamental in
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Consider the function f(x) = 5/4x^4 + 2/3x^3 - 5x^2 - 4x + 5
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y = 5/4x^4 + 2/3x^3 - 5x^2 - 4x + 5 y' = 5x^3 + 2x^2 - 10x - 4 = (5x+2)(x^2-2) y" = 15x^2 + 4x - 10
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Find the functional values requested in the problem.
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Incomplete.
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