If #p( a ) = 6a ^ { 3} - 20a - 10#, what is #p ( - 6) and p ( \frac { 1} { 6} )#?

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Answer 1 · 2017-12-02T01:18:00

#p(-6)=-1196#

#p(\frac{1}{6})=-\frac{479}{36}# or #-13\frac{11}{36}#

To find the values of #p(-6)# and #p(1/6)#, we just need to plug those values into the function #6a^3-20a-10# for #a#.

Let’s plug in #-6# for #p# first:

#p(a)=6a^3-20a-10#

#\implies p(-6)=6(-6)^3-20(-6)-10#

#\implies p(-6)=6(-216)-(-120)-10#

#\implies p(-6)=-1296+120-10#

#\implies p(-6)=-1176-10#

#\implies p(-6)=-1186#

Now we do the same for #p=\frac{1}{6}#:

#p(a)=6a^3-20a-10#

#\implies p(\frac{1}{6})=6(\frac{1}{6})^3-20(\frac{1}{6})-10#

#\implies p(\frac{1}{6})=6(\frac{1}{216})-\frac{10}{3}-10#

#\implies p(\frac{1}{6})=\frac{1}{36}-\frac{10}{3}-10#

#\implies p(\frac{1}{6})=\frac{2}{72}-\frac{240}{72}-\frac{720}{72}#

#\implies p(\frac{1}{6})=-\frac{958}{72}#

That can also be expressed as #-13\frac{11}{33}#