Evaluate and simplify f(b/3)

Jan 2016
7
0
USA
assume that f(x)= x+7/x^2+4 for every real number x. evaluate and simplify f(b/3).
 

Plato

MHF Helper
Aug 2006
22,507
8,664
assume that f(x)= x+7/x^2+4 for every real number x. evaluate and simplify f(b/3).
This is basic simple algebra: $\displaystyle{\frac{{\frac{b}{3} + 7}}{{{{\left( {\frac{b}{3}} \right)}^2} + 4}}}$
 
  • Like
Reactions: 1 person
Jan 2016
7
0
USA
Thanks for the response. That is what I got originally. Unfortunately, the online platform I'm entering the answer into is marking it wrong. I guess I'll have to play around with it for a little bit. Thanks again.
 

Plato

MHF Helper
Aug 2006
22,507
8,664
Thanks for the response. That is what I got originally. Unfortunately, the online platform I'm entering the answer into is marking it wrong. I guess I'll have to play around with it for a little bit. Thanks again.
The OP does contain some ambiguity. Review the question to be sure if you need to add grouping symbols.
 
Feb 2014
1,748
651
United States
If the question asked you to find and simplify, then Plato was leaving you to do the simplification. The computer may not accept anything but a fully simplified answer, which is NOT

$\dfrac{\dfrac{b}{3} + 7}{\left ( \dfrac{b}{3} \right )^2 + 4}.$
 
Jan 2016
7
0
USA
so is the simplified version of Plato's answer 7/(b/3)+4?
 
Feb 2014
1,748
651
United States
so is the simplified version of Plato's answer 7/(b/3)+4?
No, not at all, and please use grouping symbols properly. You mean 7/{(b/3) + 4}. You need to review PEMDAS.

Here is your error

$\dfrac{\dfrac{b}{3} + 7}{\left ( \dfrac{b}{3} \right )^2 + 4}$ $\ne$ $\dfrac{\cancel {\dfrac{b}{3}} + 7}{\left ( \cancel {\dfrac{b}{3}} * \dfrac{b}{3} \right ) + 4} = \dfrac{7}{\dfrac{b}{3} + 4}.$ THIS IS WRONG.

For example

$\dfrac{3 + 12}{3 + 2} = \dfrac{15}{5} = 3\ NOT\ \dfrac{\cancel 3 + 12}{\cancel 3 + 2} = \dfrac{12}{2} = 6.$

You must turn both numerator and denominator into complete fractions, and then invert the denominator to multiply by the numerator.

$\dfrac{\dfrac{b}{3} + 7}{\left ( \dfrac{b}{3} \right )^2 + 4} = \dfrac{\dfrac{b}{3} + \dfrac{21}{3}}{\dfrac{b^2}{9} + 4} =$

$\dfrac{\dfrac{b + 21}{3}}{\dfrac{b^2}{9} + \dfrac{36}{9}} = \dfrac{\dfrac{b + 21}{3}}{\dfrac{b^2 + 36}{3 * 3}} =\dfrac{b + 21}{\cancel 3} * \dfrac{\cancel 3 * 3}{b^2 + 36} = \dfrac{3(b + 21)}{b^2 + 36}.$
 
  • Like
Reactions: 2 people