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anonymous

  • one year ago

A graph of 2 functions is shown below. graph of function f of x equals negative 2 multiplied by x plus 3 and graph of function g of x equals x cubed plus 4 multiplied by x squared minus x minus 4 Which of the following is an approximate solution for f(x) = g(x)? x = 1 x = −1 x = −4 x = 2

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  1. anonymous
    • one year ago
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    @Vocaloid

  2. anonymous
    • one year ago
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    @ospreytriple

  3. anonymous
    • one year ago
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    @Michele_Laino

  4. anonymous
    • one year ago
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    @LiveLaughDie

  5. Plasmataco
    • one year ago
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    Set the 2 equations to equal each other

  6. Michele_Laino
    • one year ago
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    the equations of your functions, are: \[\Large \begin{gathered} f\left( x \right) = - 2x + 3 \hfill \\ g\left( x \right) = {x^3} + 4{x^2} - 4 \hfill \\ \end{gathered} \] now, you have to establish, what vauel of x, among that you have listed above, is the one such that g(x)=f(x)

  7. Michele_Laino
    • one year ago
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    value*

  8. Plasmataco
    • one year ago
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    U get 2x+3=X powered 3+4x powered 2-x-4

  9. anonymous
    • one year ago
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    okay :]

  10. Plasmataco
    • one year ago
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    Simplify it so that 0 is on one side

  11. Michele_Laino
    • one year ago
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    for example, let's consider x=-1, then we have: \[\Large \begin{gathered} f\left( { - 1} \right) = - 2 \cdot \left( { - 1} \right) + 3 = 5 \hfill \\ g\left( { - 1} \right) = {\left( { - 1} \right)^3} + 4{\left( { - 1} \right)^2} - 4 = - 1 + 4 - 4 = - 1 \hfill \\ \end{gathered} \] so our value for x, can not be x=-1

  12. Plasmataco
    • one year ago
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    Now u get X powered 3+4x powered 2+x-7

  13. Plasmataco
    • one year ago
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    Which is all. Equal to 0

  14. anonymous
    • one year ago
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    okay :]

  15. Plasmataco
    • one year ago
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    Now, factor is to some thing like (x+a)(x+b)(x+c)

  16. Michele_Laino
    • one year ago
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    plese, try with x=1, namely replace x with 1, into both functions f(x), and g(x), what do you get?

  17. anonymous
    • one year ago
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    ok give me 1 min ;]

  18. Michele_Laino
    • one year ago
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    please*

  19. anonymous
    • one year ago
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    \[f(−4)=−2⋅(−4)+3=5g(−4)=(−4)^3+4(−4)^2−4=-4+4−4=-4?\]

  20. Plasmataco
    • one year ago
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    I'll let u do it @Michele_Laino s way but with this, u get an approximation of 1

  21. Michele_Laino
    • one year ago
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    we have: \[\Large \begin{gathered} f\left( { - 4} \right) = - 2\left( { - 4} \right) + 3 = 11 \hfill \\ g\left( { - 4} \right) = {\left( { - 4} \right)^3} + 4{\left( { - 4} \right)^2} - 4 = - 64 + 64 - 4 = - 4 \hfill \\ \end{gathered} \]

  22. anonymous
    • one year ago
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    Oh

  23. Michele_Laino
    • one year ago
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    please compute these quantities: \[\Large \begin{gathered} f\left( 1 \right) = ... \hfill \\ g\left( 1 \right) = ... \hfill \\ \end{gathered} \]

  24. anonymous
    • one year ago
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    So its 1?

  25. Michele_Laino
    • one year ago
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    we have: \[\Large \begin{gathered} f\left( 1 \right) = - 2 \cdot 1 + 3 = - 2 + 3 = 1 \hfill \\ g\left( 1 \right) = {1^3} + 4 \cdot {1^2} - 4 = 1 + 4 - 4 = 1 \hfill \\ \end{gathered} \]

  26. Michele_Laino
    • one year ago
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    so, it is x=1

  27. anonymous
    • one year ago
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    Thank you!

  28. Michele_Laino
    • one year ago
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    :)

  29. anonymous
    • one year ago
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    Can you check one for me?

  30. Michele_Laino
    • one year ago
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    ok!

  31. anonymous
    • one year ago
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    Wait nvm! I got it !Thank you!

  32. Michele_Laino
    • one year ago
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    ok! :)

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