Add and subtract functions defined by equations - High school Precalculus Exercises

Let f be the function defined by f\left(x\right)=2x^2+3, and let g be the function defined by g\left(x\right)=-3x^2-2x. Determine an expression of f+g.

\left(f+g\right)\left(x\right)=-5x^2-3x+2

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

\left(f+g\right)\left(x\right)=-5x^2-2x+3

\left(f+g\right)\left(x\right)=x^2+2x+3

We can write:

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

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

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

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

Assume that f\left(x\right)=x^2+3x and g\left(x\right)=-x^2+2x+4. Determine an expression of f-g.

\left(f-g\right)\left(x\right) = 2x^2 +5x+4

\left(f-g\right)\left(x\right) = -2x^2+ x-4

\left(f-g\right)\left(x\right) = 2x^2+ x-4

\left(f-g\right)\left(x\right) = x-4

\left(f-g\right)\left(x\right) = f\left(x\right) - g\left(x\right)\\ = x^2+3x - \left(-x^2 +2x + 4\right) \\= x^2+ 3x + x^2 - 2x - 4 \\=2x^2+ x-4

\left(f-g\right)\left(x\right) = 2x^2+ x-4

Assume that f\left(x\right)= \sqrt{x^2+1} and g\left(x\right) =\dfrac{x^2-1}{ \sqrt{x^2+1}}. Determine an expression of f+g.

\left(f+g\right)\left(x\right)= \dfrac{2x^2-1}{ { \sqrt{x^2+1}}}

\left(f+g\right)\left(x\right)= \dfrac{2x^2}{ { \sqrt{x^2+1}}}

\left(f+g\right)\left(x\right)= \dfrac{x^2}{ { \sqrt{x^2+1}}}

\left(f+g\right)\left(x\right)= \dfrac{x^2+ 1}{ { \sqrt{x^2+1}}}

\left(f+g\right)\left(x\right) = f\left(x\right) + g\left(x\right)\\ = { \sqrt{x^2+1}}+ \dfrac{x^2-1}{ \sqrt{x^2+1}} \\= \dfrac{ { \sqrt{x^2+1}}^2+x^2-1}{ { \sqrt{x^2+1}}} \\= \dfrac{x^2+1+x^2-1}{ { \sqrt{x^2+1}}}\\=\dfrac{2x^2}{ { \sqrt{x^2+1}}}

\left(f+g\right)\left(x\right) = \dfrac{2x^2}{ { \sqrt{x^2+1}}}

Assume that f\left(x\right) = \dfrac{2}{x+1} and g\left(x\right) = \dfrac{x-1}{x+1}. Determine an expression of f+g.

\left(f+g\right)\left(x\right)= 1

\left(f+g\right)\left(x\right)= \dfrac{x-1}{x+1}

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

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

\left(f+g\right)\left(x\right) = f\left(x\right) + g\left(x\right)\\= \dfrac{2}{x+1} + \dfrac{x-1}{x+1} \\=\dfrac{2+x-1}{x+1}\\=\dfrac{x+1}{x+1} \\= \dfrac{x+1}{x+1}=1

\left(f+g\right)\left(x\right) = 1

Assume that f\left(x\right) = 2x and g\left(x\right) = \dfrac{x}{x+1}. Determine an expression of f+g.

\left(f-g\right)\left(x\right)= \dfrac{2x^2-x}{x+1}

\left(f-g\right)\left(x\right)= \dfrac{2x^2+x}{x+1}

\left(f-g\right)\left(x\right)= \dfrac{x^2+x}{x+1}

\left(f-g\right)\left(x\right)= \dfrac{x^2-x}{x+1}

\left(f-g\right)\left(x\right) = f\left(x\right) - g\left(x\right) \\= 2x - \dfrac{x}{x+1} \\= \dfrac{2x^2+2x - x}{x+1}\\= \dfrac{x^2+x}{x+1}

\left(f-g\right)\left(x\right) = \dfrac{x^2+x}{x+1}

Suppose that f\left(x\right) = x^2+2x-1 and g\left(x\right) = \left(x+1\right)^2. Determine an expression of f-g.

\left(f-g\right)\left(x\right)= 2x^2+4x

\left(f-g\right)\left(x\right)= 2x^2-4x

\left(f-g\right)\left(x\right)= 2x-1

\left(f-g\right)\left(x\right)=-2

f-g\left(x\right) = f\left(x\right) - g\left(x\right)\\= x^2+2x-1 -\left(x+1\right)^2 \\= x^2+2x-1 - \left(x^2 + 2x +1\right) \\= x^2+2x-1 - x^2 - 2x -1 = \\-2

f-g\left(x\right)=-2

Suppose that f\left(x\right) = \sqrt{x}-1 and g\left(x\right) = \dfrac{x}{\sqrt{x}+1}. Determine an expression of f-g.

\left(f-g\right)\left(x\right)= \dfrac{x}{\sqrt{x}+1}

\left(f-g\right)\left(x\right)= \dfrac{-1}{\sqrt{x}+1}

\left(f-g\right)\left(x\right)= \dfrac{1}{\sqrt{x}+1}

\left(f-g\right)\left(x\right)= \dfrac{-x}{\sqrt{x}+1}

\left(f-g\right)\left(x\right) = f\left(x\right) - g\left(x\right)\\ = { \sqrt{x}}- 1- \dfrac{x}{ \sqrt{x}+1} \\= \dfrac{ { {x-1-x}}}{ { \sqrt{x}+1}} \\= \dfrac{-1}{ { \sqrt{x}+1}}\\

\left(f-g\right)\left(x\right) = \dfrac{-1}{\sqrt{x}+1}