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anonymous

  • 5 years ago

Determine the equation of the line tangent to the graph of the following function at x = 0. g(x) = e^x(−4 + 3x + 3x^2)

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  1. bahrom7893
    • 5 years ago
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    okay tangent line is the first derivative

  2. bahrom7893
    • 5 years ago
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    g(x) = e^x(−4 + 3x + 3x^2) g'(x) = (3+6x) * e^x(−4 + 3x + 3x^2)

  3. bahrom7893
    • 5 years ago
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    woops sorry the derivative was wrong..

  4. bahrom7893
    • 5 years ago
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    didn't see the x, btw is it (e^x)(−4 + 3x + 3x^2)?

  5. bahrom7893
    • 5 years ago
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    i mean is e^x separate?

  6. anonymous
    • 5 years ago
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    product rule?

  7. bahrom7893
    • 5 years ago
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    well yeah, but he doesn't specify... People use the freakin parenthesis, we have to guess what the heck was the original problem... Is (−4 + 3x + 3x^2) in the power too?

  8. bahrom7893
    • 5 years ago
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    i mean we are all busy students and u guys can't even make our lives easier?? Please, use a calculator notation, we will understand it, sorry if I sounded to harsh earlier, I was mad at someone else.. =)

  9. anonymous
    • 5 years ago
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    I think the equation of the tangent line is Y=-x-4

  10. anonymous
    • 5 years ago
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    first input the value of x as g(0) to fine the value of y. Then find the derivative using the product rule. factor out the e^x, and collect like terms. Then input the value of x into the derivative to find the slope. g'(x)=m. then use point slope formula.

  11. anonymous
    • 5 years ago
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    dear hix212, thank you for the help it was -x-4

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