Question

Find (f∙g)(x) (hint: After you multiply them, factor both the numerator and denominator. Is there anything that cancels out?): f(x)=2/(x^2-4) g(x)= x^2+4x+4 2. What is the domain and range of (f * g)(x) from number 1? (Hint: Find the domain of f(x) and g(x) first, then look at (f*g)(x) and put them all together.

Answer 1

Answer:

The domain of $(f\cdot g)(x)$ is $R-\{-2,2\}$.

Step-by-step explanation:

Given that,

$f(x)=\frac{2}{x^2-4}$ and

$g(x)=x^2+4x+4.$

The function f(x) is not defined when the denominator equals zero, i.e

$x^2-4=0$

$\Rightarrow x=\pm2$

So, the domain of f(x) is all real number except $\pm2$.

$f(x)=\frac{2}{x^2-4}$

for $x\in R-\{-2,2\}\;\cdots(i)$

The function g(x) is defined for all real numbers.

$g(x)=x^2+4x+4$

for $x\in R\;\cdots(ii)$

As $(f\cdot g)(x)=f(x)\cdot g(x)$

$\Rightarrow(f\cdot g)(x)=\left(\frac{2}{x^2-4}\right)\cdot \left(x^2+4x+4\right)$

$\Rightarrow \frac{2(x^2+4x+4)}{x^2-4}$

this resulting function have the common domain of f(x) and g(x).

So, from equation (i) and (ii), the domain of $(f\cdot g)(x)$ is all real numbers except $\pm2$. i.e

$(f\cdot g)(x)=\frac{2(x^2+4x+4)}{x^2-4}$  for $x\in R-\{-2,2\}$