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sesenic [268]
1 year ago
9

See attached picture, is there a way to enter equations or algebraic symbols?

Mathematics
1 answer:
coldgirl [10]1 year ago
5 0

SOLUTIONS

Solve a given equations or algebraic symbols?

\begin{gathered} f(x)=x^2+2x \\ g(x)=1-x^2 \end{gathered}

(A)

\begin{gathered} (f+g)(x)=(x^2+2x)+(1-x^2) \\ collect\text{ like terms} \\ x^2-x^2+2x+1 \\ (f+g)(x^)=2x+1 \end{gathered}

(B)

\begin{gathered} (f-g)(x)=(x^2+2x)-(1-x^2) \\ =x^2+2x-1+x^2 \\ =x^2+x^2+2x-1 \\ (f-g)(x)=2x^2+2x-1 \end{gathered}

(C)

\begin{gathered} fg(x)=(x^2+2x)(1-x^2) \\ =x^2-x^4+2x-2x^3 \\ fg(x)=-x^4-2x^3+x^2+2x \end{gathered}

(D)

\begin{gathered} \frac{f}{g}(x)=(x^2+2x)\div(1-x^2) \\ =\frac{x^2+2x}{1-x^2} \end{gathered}

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So we have the two functions:

f(x)=2x^2+2x-3\text{ and } g(x)=x-3

And we want to find (f+g)(x), (f-g)(x), and (f*g)(x).

1)

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2)

(f-g)(x) is the same to f(x)-g(x). So:

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Distribute:

=(2x^2+2x-3)+(-x+3)

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=(2x^2)+(2x-x)+(-3+3)

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3)

(f*g)(x) is the same to f(x)*g(x). Thus:

(f\cdot g)(x)=f(x)\cdot g(x)\\=(2x^2+2x-3)(x-3)

Distribute:

=(2x^2+2x-3)(x)+(2x^2+2x-3)(-3)

Distribute:

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Combine like terms:

=(2x^3)+(2x^2-6x^2)+(-3x-6x)+(9)

Simplify:

=2x^3-4x^2-9x+9

So:

(f \cdot g)(x)=2x^3-4x^2-9x+9

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