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Math Help - (6)'=0

  1. #1
    Boo
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    (6)'=0

    Please!
    Stupid q, but...
    What is the proof that (5)' =0?

    f(x)=5
    f'(x)=0

    I see that the graph has NO SLOPE, but how to proove it arithmetically?

    Many thanks!
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  2. #2
    MHF Contributor
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    Re: (6)'=0

    5 is a constant. The derivative of any constant is 0.

    If you want a more rigorous proof:

    \displaystyle \begin{align*} f(x) &= 5 \\ &= 5x^0 \\ \\ f'(x) &= 0\cdot 5 x^{0 - 1} \\ &= 0x^{-1} \\ &= 0 \end{align*}

    Or an even more rigorous proof:

    \displaystyle \begin{align*} f(x) &= 5 \\ \\ f(x + h) &= 5 \\ \\ f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{5 - 5}{h} \\ &= \lim_{h \to 0} \frac{0}{h} \\ &= \lim_{h \to 0} 0 \\ &= 0 \end{align*}
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  3. #3
    Boo
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    Re: (6)'=0

    Ingenious!!!
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  4. #4
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    Re: (6)'=0

    Or, as you pointed out to begin with, this is a "linear function" and its derivative is just a constant- its slope. If f(x)= 5, then f(1)= 5 and f(2)= 5. The slope is \frac{f(2)- f(1)}{2-1}= \frac{0}{1}= 0.
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