For $f (x) = \sqrt{x}$ $f(x) = \sqrt x$ and $g (x) = x - 1$ $g(x) = x- 1$ , what is $(f \cdot g) (x); (g \cdot f) (x); (f \cdot g) (2) and (g \cdot f) (2)$ $(f * g)(x); (g * f)(x); (f * g) (2) and (g * f)(2)$ ?

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For $f (x) = \sqrt{x}$ $f(x) = \sqrt x$ and $g (x) = x - 1$ $g(x) = x- 1$ , what is $(f \cdot g) (x); (g \cdot f) (x); (f \cdot g) (2) and (g \cdot f) (2)$ $(f * g)(x); (g * f)(x); (f * g) (2) and (g * f)(2)$ ?

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Answer 1 · 2017-10-19T01:46:46

$\sqrt{x - 1}$ $sqrt(x-1)$ ; $\sqrt{x} - 1$ $sqrt(x)-1$ ; $1$ $1$ ; $0.41$ $0.41$

Explanation:

$(f \cdot g) (x)$ $(f*g)(x)$ can be rewritten as $f (g (x))$ $f(g(x))$
This means that for an x-value, you input that into $g (x)$ $g(x)$ , and then input your solution for that into $f (x)$ $f(x)$ .
$f (g (2)) = \sqrt{2 - 1}$ $f(g(2))=sqrt(2-1)$
$\sqrt{1} = 1$ $sqrt(1)=1$
$(g \cdot f) (x)$ $(g*f)(x)$ can be rewritten as $g (f (x)) .$ $g(f(x)).$ This time we start with $f (x)$ $f(x)$ and plug that into $g (x) .$ $g(x).$
$g (f (2)) = (\sqrt{2}) - 1$ $g(f(2))=(sqrt(2))-1$ (Its good to memorize that $\sqrt{2} \approx 1.41$ $sqrt(2)~~1.41$ )
$1.41 - 1$ $1.41-1$
$2$ $2$

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For $f (x) = \sqrt{x}$ $f(x) = \sqrt x$ and $g (x) = x - 1$ $g(x) = x- 1$ , what is $(f \cdot g) (x); (g \cdot f) (x); (f \cdot g) (2) and (g \cdot f) (2)$ $(f * g)(x); (g * f)(x); (f * g) (2) and (g * f)(2)$ ?

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For f(x)=√xf(x) = \sqrt x and g(x)=x−1g(x) = x- 1, what is (f⋅g)(x);(g⋅f)(x);(f⋅g)(2)and(g⋅f)(2)(f * g)(x); (g * f)(x); (f * g) (2) and (g * f)(2)?

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For $f (x) = \sqrt{x}$ $f(x) = \sqrt x$ and $g (x) = x - 1$ $g(x) = x- 1$ , what is $(f \cdot g) (x); (g \cdot f) (x); (f \cdot g) (2) and (g \cdot f) (2)$ $(f * g)(x); (g * f)(x); (f * g) (2) and (g * f)(2)$ ?