Respuesta :

r3t40

[tex]

f(x)=x+1 \\

g(x)=\dfrac{1}{x} \\

(f\cdot g)(x)=(x+1)\dfrac{1}{x} \\

(f\cdot g)(x)=\underline{\dfrac{x+1}{x}} \\ \\

0=\dfrac{x+1}{x} \\

0=\dfrac{x}{x}+\dfrac{1}{x} \\

0=1+\dfrac{1}{x} \\

-1=\dfrac{1}{x} \\

-x=1 \\

x=1

[/tex]

kudzordzifrancis

ANSWER

[tex]y \ne1[/tex]

EXPLANATION

The given functions are

[tex]f(x) = x + 1[/tex]

and

[tex]g(x) = \frac{1}{x} [/tex]

We want to find

[tex](f \times g)(x)[/tex]

We use function properties to obtain:

[tex](f \times g)(x) = f(x) \times g(x)[/tex]

[tex](f \times g)(x) = (x + 1) \times \frac{1}{x} = \frac{x + 1}{x} [/tex]

There is a horizontal asymptote at:

[tex]y = 1[/tex]

Let

[tex]y = \frac{x + 1}{x} [/tex]

[tex]xy = x + 1[/tex]

[tex]xy - x = 1[/tex]

[tex]x(y - 1) = 1[/tex]

[tex]x = \frac{1}{y - 1} [/tex]

The range is

[tex]y \ne1[/tex]

Or

[tex]( - \infty ,1) \cup(1, \infty )[/tex]

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