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qtexpress101

  • 3 years ago

find f ' (x) and f '(c) f(x)= (1/3)(2x^3 - 4) value of c = 0

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  1. qtexpress101
    • 3 years ago
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    i'm not really sure how to get started

  2. across
    • 3 years ago
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    \[ f(x)=\frac13\left(2x^3-4\right)\\ f'(x)=2x^2\\ f'(0)=0. \]

  3. qtexpress101
    • 3 years ago
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    how did you get those primes?

  4. across
    • 3 years ago
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    I used both\[ f(x)=ax^{b}\Longrightarrow f'(x)=abx^{b-1}, \]and\[ f(x)=c\Longrightarrow f'(x)=0. \]

  5. qtexpress101
    • 3 years ago
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    ah ok, thanks

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