Piecewise Function Coefficients
Suppose that $ a $ and $ b $ are positive real numbers, and let \[f(x) =
\begin{cases}
\frac{a}{b}x & \text{ if }x\le-4, \\
abx^2 & \text{ if }x>-4.\end{cases}
\]If $ f(-4)=-\frac{60}{13} $ and $ f(4)=3120 $, what is $ a+b $?
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- $\frac{a}{b}$
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- 0
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- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
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- $\infty$
Solution
Since $ -4\le-4 $, we know that $ f(-4)=\frac{a}{b}(-4)=-\frac{60}{13} $. So, $ \frac{a}{b}=\frac{15}{13} $. Next we look at $ 4>-4 $, so $ f(4)=ab\cdot4^2=3120 $. That means $ ab=\frac{3120}{16}=195 $. Now that we have two equations and two variables, we can solve for $ a $ and $ b $. From $ ab=195 $, we get that $ a=\frac{195}{b} $. We substitute this value for $ a $ into the equation $ \frac{a}{b}=\frac{15}{13} $ to get $ \frac{195}{b^2}=\frac{15}{13} $.
Next we cross-multiply and get $ 15b^2=13\cdot195 $. Before multiplying 13 by 195, we try factoring 195 and notice that 15 is a factor of 195, so we can rewrite this as $ 15b^2=13\cdot13\cdot15 $. Finally, $ b^2=13^2 $, so $ b=\pm13 $. The problem says that $ a $ and $ b $ are positive, so $ b=13 $ and $ a=\frac{195}{13}=15 $. The sum $ a+b $ equals $ \boxed{28} $.
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