Question

Let f(x)= 1/x and g(x)=x^2+5x.

a. Find (f*g)(x)
b. Find the domain and range of (f*g)(x)

Answer 1

$f(x) = \frac{1}{x} \ \ \ g(x) = x^2 + 5x\\f \cdot g (x) = \frac{1}{x} \cdot (x^2 + 5x) = x + 5 => D(f \cdot g) = \Bbb R \ \ \ R(f \cdot g) = \Bbb R$

Answer 2

a) The expression is $(f\,\circ\,g)(x) = \frac{1}{x\cdot (x+5)}$.

b) The domain consists in the set of x-values so that y-values exist. Thus, we find that domain of $(f\,\circ \,g)(x)$ is $Dom ((f\,\circ\,g)(x)) = \mathbb{R}-\{0,-5\}$.

The range consists in the set of y-values of the function. Thus, we find that range of $(f\,\circ\,g)(x)$ is $Ran((f\,\circ\,g)(x)) = \mathbb{R}$.

How to analyze the composition between function

a) Let be two functions $f(x)$ and $g(x)$. There is a composition between the functions when independent variable of the former is replaced by the latter one:

$(f\,\circ \,g)(x)=\frac{1}{x^{2}+5\cdot x}$

$(f\,\circ\,g)(x) = \frac{1}{x\cdot (x+5)}$ (1)

The expression is $(f\,\circ\,g)(x) = \frac{1}{x\cdot (x+5)}$. $\blacksquare$

b) The domain consists in the set of x-values so that y-values exist. Thus, we find that domain of $(f\,\circ \,g)(x)$ is $Dom ((f\,\circ\,g)(x)) = \mathbb{R}-\{0,-5\}$. $\blacksquare$

The range consists in the set of y-values of the function. Thus, we find that range of $(f\,\circ\,g)(x)$ is $Ran((f\,\circ\,g)(x)) = \mathbb{R}$. $\blacksquare$

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