3 years ago

2

Mathematics

1 answer:

Valentin [98]3 years ago

4 0

Answer:

15 to 21

Step-by-step explanation:

junk to total

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Given the following functions below,

$\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}$

Factorising the denominators of both functions,

Factorising the denominator of f(x),

$\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}$

Factorising the denominator of g(x),

$\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}$

Multiplying both functions,

$undefined$

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1 year ago