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${:[f(x)=(1)/(2x^(2)+1)],[g(x)=x^(2)+5]:} Find f*g and f-g. Then, give their don (f*g)(x)= Domain of f*g : (f-g)(x)=$

$\begin{array}{l} f(x)=\frac{1}{2 x^{2}+1} \\ g(x)=x^{2}+5 \end{array}$ $\newline$ Find $f \cdot g$ and $f-g$ . Then, give their don $\newline$ $(f \cdot g)(x)=$ $\newline$ Domain of $f \cdot g$ : $\newline$ $(f-g)(x)=$

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Find derivatives of sine and cosine functions

Full solution

Q.

\begin{array}{l} f(x)=\frac{1}{2 x^{2}+1} \\ g(x)=x^{2}+5 \end{array}

\newline

Find

f \cdot g

and

f-g

. Then, give their don

\newline

(f \cdot g)(x)=

\newline

Domain of

f \cdot g

:

\newline

(f-g)(x)=

Define functions $f(x)$ and $g(x)$ : Step $1$ : Define the functions $f(x)$ and $g(x)$ . $f(x) = \frac{1}{2x^2 + 1}$ , $g(x) = x^2 + 5$
Find product $(f*g)(x)$ : Step $2$ : Find the product $(f*g)(x)$ . $(f*g)(x) = f(x) * g(x) = \left(\frac{1}{2x^2 + 1}\right) * (x^2 + 5)$
Simplify expression for $(f*g)(x)$ : Step $3$ : Simplify the expression for $(f*g)(x)$ . $(f*g)(x) = \frac{x^2 + 5}{2x^2 + 1}$
Determine domain of $(f*g)(x)$ : Step $4$ : Determine the domain of $(f*g)(x)$ . The denominator $2x^2 + 1$ is never zero, so the domain is all real numbers.
Find difference $(f-g)(x)$ : Step $5$ : Find the difference $(f-g)(x)$ . $(f-g)(x) = f(x) - g(x) = \left(\frac{1}{2x^2 + 1}\right) - (x^2 + 5)$
Simplify expression for $(f-g)(x)$ : Step $6$ : Simplify the expression for $(f-g)(x)$ . $(f-g)(x) = \frac{1 - (x^2 + 5)(2x^2 + 1)}{2x^2 + 1}$

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