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darkmare

  • 3 years ago

f(x) = -2x^2+4x for x <0 and 8x^2-3 for x greater or equal to zero what would the difference quotient's be for lim x->0- and lim x-> 0+?

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  1. ash2326
    • 3 years ago
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    Function is defined as: \[f(x)=-2x^2+4x, for\ x<0\] \[f(x)=8x^2-3. for\ x\ge 0\] for limit 0-, f(x) defined for x<0 \[\large \lim_{x \to 0^{-}} (-2x^2+4x)\] for limit 0+, f(x) defined for \(x\ge 0\) \[\large \lim_{x \to 0^{+}} (8x^2-3)\]

  2. ash2326
    • 3 years ago
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    @darkmare do you get this?

  3. darkmare
    • 3 years ago
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    yes I understand that for the limit of each one that those are used but the questions asks the lim of those values for f prime (0) so those are not the answers they are looking for apparently

  4. suneja
    • 3 years ago
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    f(x)=−2x2+4x,for x<0 f(x)=8x2−3.for x≥0 for limit 0-, f(x) defined for x<0 lim x→0−(−2x2+4x)= 0 for limit 0+, f(x) defined for x≥0 lim x→0+(8x2−3)=-3

  5. ash2326
    • 3 years ago
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    \[\large f'(x)=\lim_{h \to 0^-}\frac{f(x-h)-f(0)}{x-h-x}\] to evaluate f'(0), from left hand side ( or to check differentiability) put x=0 in this \[\large f'(x)=\lim_{h \to 0^-}\frac{-2(x-h)^2+4(x-h)-(-2x^2+4x)}{x-h-x}\] now put x=0 in this, for 0+, use f(x)=8x^2-3

  6. ash2326
    • 3 years ago
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    do you get it ? @darkmare

  7. darkmare
    • 3 years ago
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    yes I do understand that, it is still not accepting it as an answer so I will have to look into it further, perhaps simplify further

  8. ash2326
    • 3 years ago
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    yeah, you have to simplify it further.

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