If #f(x)=5x#, #g(x)=1/x#, and #h(x)=x^3#, how do you find #(5h(1)-2g(3))/(3f(2))#?

2 Answers
Jan 28, 2016

#13/90#, or #0.1bar4#

Explanation:

#(5h(1) - 2g(3))/(3f(2))#

Now let's plug in the values for those functions:

#(5*(1^3) - 2*(1/3))/(3*(5*2))#

#1^3 = 1#, and since any number times #1# is the same, we can remove it.

#(5 - 2*(1/3))/(3*(5*2))#

Now we can simplify #2*(1/3)# to #2/3#

#(5 - 2/3)/(3*(5*2))#

And we can multiply the numbers in the denominator #3*5 = 15#, and #15 *2 = 30#

#(5 - 2/3)/30#

Now let's try and get rid of that ugly fraction in a fraction! We can do that by multiplying the entire fraction by #3/3# which is also equal to #1#, and like we said above, we're not actually changing the value by doing that!

#(5 - 2/3)/30 * 3/3#

Multiplying through on both sides:

#(5*3 - (2/3)*3)/(30*3)#

Now simplify the multiplications:

#(15 - 2)/90#

Now subtract in the numerator... and...

#13/90#

There you have it!

Jan 28, 2016

#13/90#

Explanation:

#(5h(1)-2g(3))/(3f(2))#
#=(5(1)^3-2(1/3))/(3(5*2)#
#=(5-2/3)/(3*10)#
#=(13/3)/30#
#=13/3*1/30#
#=13/90#