Find \((f+g)(x)\), \((f-g)(x)\), \((fg)(x)\) and \((f/g)(x)\)
Question
Let
\[ f(x)=x^2 \]and
\[ g(x)=2x+1 \]be two real functions. Find
\((f+g)(x)\), \((f-g)(x)\), \((fg)(x)\) and \((f/g)(x)\).
Solution
Given
\[ f(x)=x^2 \]and
\[ g(x)=2x+1 \]Find \((f+g)(x)\)
By definition,
\[ (f+g)(x)=f(x)+g(x) \] \[ =x^2+(2x+1) \] \[ =x^2+2x+1 \]Therefore,
\[ (f+g)(x)=x^2+2x+1 \]Find \((f-g)(x)\)
By definition,
\[ (f-g)(x)=f(x)-g(x) \] \[ =x^2-(2x+1) \] \[ =x^2-2x-1 \]Therefore,
\[ (f-g)(x)=x^2-2x-1 \]Find \((fg)(x)\)
By definition,
\[ (fg)(x)=f(x)\cdot g(x) \] \[ =x^2(2x+1) \] \[ =2x^3+x^2 \]Therefore,
\[ (fg)(x)=2x^3+x^2 \]Find \((f/g)(x)\)
By definition,
\[ \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)} \] \[ =\frac{x^2}{2x+1} \]Denominator cannot be zero.
\[ 2x+1\ne0 \] \[ x\ne-\frac12 \]Therefore,
\[ \left(\frac{f}{g}\right)(x)=\frac{x^2}{2x+1}, \qquad x\ne-\frac12 \]