Let $f (x) = x^{2} + 6$ $f(x) = x^2 + 6$ and $g (x) = \frac{x + 8}{x}$ $g(x) = (x + 8)/(x)$ find $(g \cdot f) (- 7)$ $(g*f)(-7)$ ?

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Let $f (x) = x^{2} + 6$ $f(x) = x^2 + 6$ and $g (x) = \frac{x + 8}{x}$ $g(x) = (x + 8)/(x)$ find $(g \cdot f) (- 7)$ $(g*f)(-7)$ ?

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Answer 1 · 2018-03-15T22:36:28

$- \frac{55}{7}$ $-55/7$

Explanation:

Let's start by multiplying our functions, $f$ $f$ and $g$ $g$ . We are essentially setting up the following equation

$\frac{x^{2} + 6}{1} = \frac{x + 8}{x}$ $(x^2+6)/1=(x+8)/x$

Where we are essentially only multiplying the numerator to get

$\frac{(x^{2} + 6) (x + 8)}{x}$ $(color(blue)((x^2+6)(x+8)))/x$

We can use FOIL (Firsts, Outsides, Insides, Lasts) to multiply this binomial. We get

$\frac{x^{2} (x) + x^{2} (8) + 6 (x) + 6 (8)}{x}$ $(x^2(x)+x^2(8)+6(x)+6(8))/x$

Which is equal to

$\frac{x^{3} + 8 x^{2} + 6 x + 48}{x}$ $(x^3+8x^2+6x+48)/x$ as our new function. We can plug in $- 7$ $-7$ to get

$\frac{{(- 7)}^{3} + 8 {(- 7)}^{2} + 6 (- 7) + 48}{- 7}$ $((-7)^3+8(-7)^2+6(-7)+48)/-7$

Which simplifies to

$\frac{- 343 + 392 - 42 + 48}{- 7}$ $(-343+392-42+48)/-7$

$\Rightarrow - \frac{55}{7}$ $=>-55/7$

Answer 2 · 2018-03-16T07:49:02

$(g \cdot f) (- 7) = - \frac{55}{7}$ $(g * f)(-7) = -55/7$

$(g \circ f) (- 7) = \frac{63}{55}$ $(g @ f)(-7) = 63/55$

Explanation:

Given:

$f (x) = x^{2} + 6$ $f(x) = x^2+6$

$g (x) = \frac{x + 8}{x}$ $g(x) = (x+8)/x$

I suspect that the question is wanting the value of $(g \circ f) (- 7)$ $(g @ f)(-7)$ and not $(g \cdot f) (- 7)$ $(g * f)(-7)$ , but let us describe both...

Case $(g \cdot f) (- 7)$ $(g * f)(-7)$

Given two functions $g$ $g$ and $f$ $f$ of a single variable, we can define the product function $g \cdot f$ $g * f$ by:

$(g \cdot f) (x) = g (x) \cdot f (x)$ $(g * f)(x) = g(x) * f(x)" "$ for all $x$ $x$ in their common domain

With this definition:

$(g \cdot f) (- 7) = g (- 7) \cdot f (- 7)$ $(g * f)(-7) = g(-7) * f(-7)$

$(g \cdot f) (- 7) = \frac{(- 7) + 8}{- 7} \cdot ({(- 7)}^{2} + 6)$ $color(white)((g * f)(-7)) = ((-7)+8)/(-7) * ((-7)^2 + 6)$

$(g \cdot f) (- 7) = (- \frac{1}{7}) \cdot (49 + 6)$ $color(white)((g * f)(-7)) = (-1/7) * (49 + 6)$

$(g \cdot f) (- 7) = - \frac{55}{7}$ $color(white)((g * f)(-7)) = -55/7$

Case $(g \circ f) (- 7)$ $(g @ f)(-7)$

Given two functions $g$ $g$ and $f$ $f$ of a single variable, we can define their compostion $g \circ f$ $g @ f$ by:

$(g \circ f) (x) = g (f (x)) \frac{0}{0}$ $(g @ f)(x) = g(f(x))color(white)(0/0)$
for all $x$ $x$ in the domain of $f (x)$ $f(x)$ such that $f (x)$ $f(x)$ is in the domain of $g (x)$ $g(x)$

With this definition:

$(g \circ f) (- 7) = g (f (- 7))$ $(g @ f)(-7) = g(f(-7))$

$(g \circ f) (- 7) = g ({(- 7)}^{2} + 6)$ $color(white)((g @ f)(-7)) = g((-7)^2 + 6)$

$(g \circ f) (- 7) = g (49 + 6)$ $color(white)((g @ f)(-7)) = g(49 + 6)$

$(g \circ f) (- 7) = g (55)$ $color(white)((g @ f)(-7)) = g(55)$

$(g \circ f) (- 7) = \frac{(55) + 8}{(55)}$ $color(white)((g @ f)(-7)) = ((55)+8)/((55))$

$(g \circ f) (- 7) = \frac{63}{55}$ $color(white)((g @ f)(-7)) = 63/55$

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Let $f (x) = x^{2} + 6$ $f(x) = x^2 + 6$ and $g (x) = \frac{x + 8}{x}$ $g(x) = (x + 8)/(x)$ find $(g \cdot f) (- 7)$ $(g*f)(-7)$ ?

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Let f(x) = x^2 + 6f(x)=x2+6f(x) = x^2 + 6 and g(x) = (x + 8)/(x)g(x)=x+8xg(x) = (x + 8)/(x) find (g*f)(-7)(g⋅f)(−7)(g*f)(-7)?

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Let $f (x) = x^{2} + 6$ $f(x) = x^2 + 6$ and $g (x) = \frac{x + 8}{x}$ $g(x) = (x + 8)/(x)$ find $(g \cdot f) (- 7)$ $(g*f)(-7)$ ?