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interval and symmetry questions
- Thread starter xingz
- Start date
For #5:
\(\displaystyle \L\\g(x)=2x^{3}+mx^{2}+nx+7\)
We know that (2,5) is on the graph.
\(\displaystyle \L\\2(2)^{3}+m(2)^{2}+2n+7=5\Rightarrow{4m+2n=-18}\)
We need another equation in order to solve for m and n.
Dividing g(x) by x+3, we get:
\(\displaystyle \L\\2x^{2}+mx-6x-3m+n+18+\frac{9m-3n-47}{x+3}\)
The remainder is \(\displaystyle 9m-3n-47=-95\)
Now, you have 2 equation with 2 unknowns:
\(\displaystyle \L\\9m-3n=-48\)
\(\displaystyle \L\\4m+2n=-18\)
Solve for m and n.
\(\displaystyle \L\\g(x)=2x^{3}+mx^{2}+nx+7\)
We know that (2,5) is on the graph.
\(\displaystyle \L\\2(2)^{3}+m(2)^{2}+2n+7=5\Rightarrow{4m+2n=-18}\)
We need another equation in order to solve for m and n.
Dividing g(x) by x+3, we get:
\(\displaystyle \L\\2x^{2}+mx-6x-3m+n+18+\frac{9m-3n-47}{x+3}\)
The remainder is \(\displaystyle 9m-3n-47=-95\)
Now, you have 2 equation with 2 unknowns:
\(\displaystyle \L\\9m-3n=-48\)
\(\displaystyle \L\\4m+2n=-18\)
Solve for m and n.
For #3. Graph the function. Check to see if it's even or odd.
If even, then f(-x)=f(x). That is, if you put -x in place of x in the function, you get the same thing as the original.
If it's odd, then f(-x)=-f(x). That is, if you put -x in your function you get back the negative of it.
For #4b. Play around with some values and see where it's greater than 0 and where it's less than 0.
If even, then f(-x)=f(x). That is, if you put -x in place of x in the function, you get the same thing as the original.
If it's odd, then f(-x)=-f(x). That is, if you put -x in your function you get back the negative of it.
For #4b. Play around with some values and see where it's greater than 0 and where it's less than 0.
For 4b you have the x-intercepts, which divide the domain into regions where f(x) is >0 or <0, so all that is left to do is test the sign of f(x) at values of x between the x-intercepts.xingz said:Can anyone help me out with question 3 and question 4 part b? thanks