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$f(x) = \begin{cases} \frac{a}{b}x & \text{ if }x\le-4, \\ abx^2 & \text{ if }x>-4. \end{cases}$

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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 $$a+b$$?

Aug 28, 2021

#2
+114875
+1

$$f(-4)=\frac{a}{b}*-4=\frac{-4a}{b}\\ \frac{-4a}{b}=\frac{-60}{13}\\ \frac{4a}{b}=\frac{60}{13}\\ \frac{a}{b}=\frac{15}{13}\\ a=\frac{15b}{13}$$

$$f(4)=ab*4^2=16ab\\ 16ab=3120\\ ab=195\\~\\ so\\ \frac{15b}{13}*b=195\\ \frac{b^2}{13}=13\\ b^2=13^2\\ b=\pm13$$

You can check what I have done and continue from there.

LaTex:

f(-4)=\frac{a}{b}*-4=\frac{-4a}{b}\\
\frac{-4a}{b}=\frac{-60}{13}\\
\frac{4a}{b}=\frac{60}{13}\\
\frac{a}{b}=\frac{15}{13}\\
a=\frac{15b}{13}

Aug 30, 2021