+23 Let $f(a) = \frac{13+a}{3a+7}$ where $a$ is restricted to positive integers. What is the maximum value of $f(a)$?
I gave up on this problem. The solution said you can write the function as:
\[f(a) = \frac{\frac13 (3a+7)}{3a+7} + \frac{32/3}{3a+7} = \frac13 + \frac{32}{9a+21}.\]
I get where to go from the last step, but how did (13 + a)/(3a + 7) turn into ((1/3)(3a+7))/(3a+7)?
+1365 Hello!
So, first, let's review. We know that
\(f(a) = \frac{\frac13 (3a+7)}{3a+7} + \frac{32/3}{3a+7}\) . I believe you understand this part!
Now, for the 1/3 in the second part where it starts getting confusing, let's first note that \(\frac{(ax+b)}{(x+b)} = \frac{a}{1} = a\). Since the polynomial is equivalent, we can cancel them out. That's what we are doing here. Canceling out the 3a + 7, we get the remaining 1/3.
Now, for the \(\frac{32}{9a+21}\). Let's note that \(\frac{a}{b} = \frac{a}{b} \cdot \frac{c}{c}\). We want to get rid of the 3 in the numerator, so we multiplythe numerator and denominator by 3.
We get \(\frac{32/3}{3a+7} = \frac{32/3}{3a+7} \cdot \frac{3}{3} = \frac{32}{3(3a+7)} = \frac{32}{9a+21}\)
So there, that's how we got from the second step to the third step.
Good luck from here!
I hope I answered your question!
Thanks! :)
