anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous

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anonymous
  • anonymous
Plasmataco
  • Plasmataco
Set the 2 equations to equal each other
Michele_Laino
  • Michele_Laino
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)
Michele_Laino
  • Michele_Laino
value*
Plasmataco
  • Plasmataco
U get 2x+3=X powered 3+4x powered 2-x-4
anonymous
  • anonymous
okay :]
Plasmataco
  • Plasmataco
Simplify it so that 0 is on one side
Michele_Laino
  • Michele_Laino
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
Plasmataco
  • Plasmataco
Now u get X powered 3+4x powered 2+x-7
Plasmataco
  • Plasmataco
Which is all. Equal to 0
anonymous
  • anonymous
okay :]
Plasmataco
  • Plasmataco
Now, factor is to some thing like (x+a)(x+b)(x+c)
Michele_Laino
  • Michele_Laino
plese, try with x=1, namely replace x with 1, into both functions f(x), and g(x), what do you get?
anonymous
  • anonymous
ok give me 1 min ;]
Michele_Laino
  • Michele_Laino
please*
anonymous
  • anonymous
\[f(−4)=−2⋅(−4)+3=5g(−4)=(−4)^3+4(−4)^2−4=-4+4−4=-4?\]
Plasmataco
  • Plasmataco
I'll let u do it @Michele_Laino s way but with this, u get an approximation of 1
Michele_Laino
  • Michele_Laino
we have: \[\Large \begin{gathered} f\left( { - 4} \right) = - 2\left( { - 4} \right) + 3 = 11 \hfill \\ g\left( { - 4} \right) = {\left( { - 4} \right)^3} + 4{\left( { - 4} \right)^2} - 4 = - 64 + 64 - 4 = - 4 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
Oh
Michele_Laino
  • Michele_Laino
please compute these quantities: \[\Large \begin{gathered} f\left( 1 \right) = ... \hfill \\ g\left( 1 \right) = ... \hfill \\ \end{gathered} \]
anonymous
  • anonymous
So its 1?
Michele_Laino
  • Michele_Laino
we have: \[\Large \begin{gathered} f\left( 1 \right) = - 2 \cdot 1 + 3 = - 2 + 3 = 1 \hfill \\ g\left( 1 \right) = {1^3} + 4 \cdot {1^2} - 4 = 1 + 4 - 4 = 1 \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
so, it is x=1
anonymous
  • anonymous
Thank you!
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
Can you check one for me?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
Wait nvm! I got it !Thank you!
Michele_Laino
  • Michele_Laino
ok! :)

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