Question

I need help with these textbook problems #12 and #13

Answer 1

We will have the following:

12. The zeros we can see are located at:

`\begin{gathered} (2-x)(3-2x)=0 \\ \\ \Rightarrow2-x=0\Rightarrow x=2 \\ \\ and \\ \\ \Rightarrow3-2x=0\Rightarrow2x=3\Rightarrow x=(3)/(2) \end{gathered}`

`\begin{gathered} x=2 \\ \\ x=(3)/(2) \end{gathered}`

13. We determine the zeros as follows:

`\begin{gathered} f(x)=9x^2+6x+1\Rightarrow f^(\prime)(x)=18x+6=0 \\ \\ \Rightarrow18x=-6\Rightarrow x=-(1)/(3) \end{gathered}`

So, there is only one zero at:

`x=-(1)/(3)`