## Exercises - Functions and their Combinations

1. If $\displaystyle{f(x) = \frac{11x}{x-7}}$, find the following:

1. $f(2)$

${\displaystyle{f(2) = \frac{11 \cdot 2}{2-7} = -\frac{22}{5}}}$

2. $f(7)$

${\displaystyle{f(7) \textrm{ is undefined, as the denominator would be zero.}}}$

3. $f(a-2)$

${\displaystyle{f(a-2) = \frac{11(a-2)}{(a-2)-7} = \frac{11a-22}{a-9}}}$

2. If $\displaystyle{f(x) = \frac{4}{3 | x |}}$, find the following:

1. $f(1)$

${\displaystyle{f(1) = \frac{4}{3 \left| 1 \right|} = \frac{4}{3}}}$

2. $f(-1)$

${\displaystyle{f(-1) = \frac{4}{3 \left| -1 \right|} = \frac{4}{3}}}$

3. the domain of $f$

${ \begin{array}{l} \textrm{Solving$3|x| = 0$, we see the denominator is zero when$x=0$}\\ \textrm{so the domain of$f$is } \{ x \in \mathbb{R} \mid x \neq 0 \} \textrm{ or, in interval notation: } (-\infty, 0) \cup (0,\infty) \end{array}}$

3. Let $f(x)=x^2-4$. Find and simplify the following:

1. $\displaystyle{f(3)}$

${\displaystyle{f(3) = 3^2 - 4 = 5}}$

2. $\displaystyle{f(2c)}$

${ \displaystyle{f(2c) = (2c)^2 - 4 = 4c^2-4}}$

3. $\displaystyle{f(x+1)}$

${\displaystyle{ \begin{array}{rcl} f(x+1) &=& (x+1)^2 - 4\\ &=& (x^2 + 2x + 1) - 4\\ &=& x^2+2x-3 \end{array}}}$

4. $\displaystyle{f(x+\Delta x)}$

${ \begin{array}{rcl} f(x + \Delta x) &=& (x+\Delta x)^2 - 4\\ &=& x^2+2x (\Delta x) + (\Delta x)^2 - 4 \end{array}}$

5. $\displaystyle{\frac{f(x)-f(-1)}{x+1}}$ (assume $x \neq -1$)

${ \begin{array}{rcl} \displaystyle{\frac{f(x)-f(-1)}{x+1}} &=& \displaystyle{\frac{(x^2-4)-((-1)^2-4)}{x+1}}\\\\ &=& \displaystyle{\frac{x^2-4-(-3)}{x+1}}\\\\ &=& \displaystyle{\frac{x^2-1}{x+1}}\\\\ &=& \displaystyle{\frac{(x-1)(x+1)}{x+1}}\\\\ &=& x-1 \end{array}}$

6. $\displaystyle{\frac{f(x+\Delta x) - f(x)}{\Delta x}}$ (assume $\Delta x \neq 0$)

${ \begin{array}{rcl} \displaystyle{\frac{f(x+\Delta x) - f(x)}{\Delta x}} &=& \displaystyle{\frac{((x+\Delta x)^2 - 4) - (x^2 - 4)}{\Delta x}}\\\\ &=& \displaystyle{\frac{(x^2 + 2x \Delta x + (\Delta x)^2 - 4) - x^2 + 4}{\Delta x}}\\\\ &=& \displaystyle{\frac{2x\Delta x + (\Delta x)^2}{\Delta x}}\\\\ &=& \displaystyle{\frac{\Delta x (2x + \Delta x)}{\Delta x}}\\\\ &=& \displaystyle{2x + \Delta x} \end{array}}$

4. For each of the following, find and simplify $\displaystyle{\frac{f(x+h)-f(x)}{h}}$     (You may assume that $h \neq 0$)

1. $\displaystyle{f(x) = 7x-6}$

${ \begin{array}{rcl} \displaystyle{\frac{f(x+h)-f(x)}{h}} &=& \displaystyle{\frac{(7(x+h)-6) - (7x-6)}{h}}\\\\ &=& \displaystyle{\frac{7x+7h-6-7x+6}{h}}\\\\ &=& \displaystyle{\frac{7h}{h}}\\\\ &=& 7 \end{array} }$

2. $\displaystyle{f(x) = 4x^2 - x}$

${ \begin{array}{rcl} \displaystyle{\frac{f(x+h)-f(x)}{h}} &=& \displaystyle{\frac{(4(x+h)^2 - (x+h)) - (4x^2 - x)}{h}}\\\\ &=& \displaystyle{\frac{4(x^2+2xh+h^2) - (x+h) - 4x^2 + x}{h}}\\\\ &=& \displaystyle{\frac{4x^2 + 8xh + 4h^2 - x - h - 4x^2 + x}{h}}\\\\ &=& \displaystyle{\frac{8xh + 4h^2 - h}{h}}\\\\ &=& \displaystyle{\frac{h(8x+4h-1)}{h}}\\\\ &=& \displaystyle{8x+4h-1} \end{array}}$

3. $\displaystyle{f(x) = 5x^3 - 2x^2}$

${ \begin{array}{rcl} \displaystyle{\frac{f(x+h)-f(x)}{h}} &=& \displaystyle{\frac{(5(x+h)^3 - 2(x+h)^2) - (5x^3 - 2x^2)}{h}}\\\\ &=& \displaystyle{\frac{5(x^3 + 3x^2h + 3xh^2 + h^3) - 2(x^2 + 2xh + h^2) - 5x^3 + 2x^2}{h}}\\\\ &=& \displaystyle{\frac{5x^3 + 15x^2h + 15xh^2 + 5h^3 - 2x^2 - 4xh - 2h^2 - 5x^3 + 2x^2}{h}}\\\\ &=& \displaystyle{\frac{15x^2h + 15xh^2 + 5h^3 - 4xh - 2h^2}{h}}\\\\ &=& \displaystyle{\frac{h(15x^2 + 15xh + 5h^2 - 4x - 2h)}{h}}\\\\ &=& \displaystyle{15x^2 + 15xh + 5h^2 - 4x - 2h} \end{array}}$

5. Find the domain of the following functions:

1. $\displaystyle{f(x) = \frac{5x}{x^2-9}}$

${ \begin{array}{l} \textrm{Noting that the denominator must not be zero, we solve } x^2 - 9 = 0\\ \textrm{Factoring, we find } (x-3)(x+3) = 0, \textrm{ requiring } x = \pm 3\\ \textrm{Thus, the domain is given by } \{ x \in \mathbb{R} \mid x \neq \pm 3 \}\\ \textrm{(or in interval notation:$(-\infty,-3) \cup (-3,3) \cup (3,\infty)$)} \end{array}}$

2. $\displaystyle{f(x) = \sqrt{6-x}}$

${ \begin{array}{l} \textrm{Noting that the expression under the radical must be non-negative,}\\ \textrm{we solve$6-x \ge 0$to discover the domain:$\{ x \in \mathbb{R} \mid x \le 6 \}$}\\ \textrm{(or in interval notation:$(-\infty,6]$).} \end{array}}$

3. $\displaystyle{g(x) = x^2 - 2 \left| x \right| + \frac{1}{3} + 7x^3}$

${\displaystyle{\textrm{all reals, } \mathbb{R}}}$

4. $\displaystyle{h(x) = \frac{7x}{2x^2 - 5x - 3}}$

${ \begin{array}{l} \textrm{Noting that the denominator must not be zero, we solve } 2x^2 - 5x - 3 = 0\\ \textrm{Factoring, we find } (2x + 1)(x - 3) = 0, \textrm{ requiring } x = 3, -1/2\\ \textrm{Thus, the domain is given by } \{ x \in \mathbb{R} \mid x \neq 3, -1/2 \}\\ \textrm{(or in interval notation:$(-\infty,-1/2) \cup (-1/2,3) \cup (3,\infty)$)} \end{array}}$

5. $\displaystyle{g(x) = \sqrt{2x-15}}$

${ \begin{array}{l} \textrm{Noting that the expression under the radical must be non-negative,}\\ \textrm{we solve$2x-15 \ge 0$to discover the domain:$\{ x \in \mathbb{R} \mid x \ge 15/2 \}$}\\ \textrm{(or in interval notation:$[15/2,\infty)$).} \end{array}}$

6. $\displaystyle{f(x) = \frac{3-4x}{x^2 + x - 6}}$

${ \begin{array}{l} \textrm{Noting that the denominator must not be zero, we solve } x^2 + x - 6 = 0\\ \textrm{Factoring, we find } (x+3)(x-2) = 0, \textrm{ requiring } x = -3, 2\\ \textrm{Thus, the domain is given by } \{ x \in \mathbb{R} \mid x \neq -3, 2 \}\\ \textrm{(or in interval notation:$(-\infty,-3) \cup (-3,2) \cup (2,\infty)$)} \end{array}}$

7. $\displaystyle{g(x) = \frac{3x^3 + 2}{\sqrt{x-4}}}$

${ \begin{array}{l} \textrm{Noting that both the expression under the radical must be non-negative, and}\\ \textrm{the denominator must not be zero, we determine$x-4$must be strictly positive.}\\ \textrm{Thus, we solve$x-4 \gt 0$to discover the domain:$\{ x \in \mathbb{R} \mid x \gt 4 \}$}\\ \textrm{(or in interval notation:$(4,\infty)$).} \end{array}}$

8. $\displaystyle{f(x) = 4x^3 + 5x^2 - \frac{2}{3} x + \sqrt{7}}$

${\displaystyle{\textrm{all reals, }\mathbb{R}}}$

9. $\displaystyle{q(x) = \frac{4}{x^2+3x-4}}$

${ \begin{array}{l} \textrm{Noting that the denominator must not be zero, we solve } x^2+3x-4 = 0\\ \textrm{Factoring, we find } (x+4)(x-1) = 0, \textrm{ requiring } x = -4,1\\ \textrm{Thus, the domain is given by } \{ x \in \mathbb{R} \mid x \neq -4,1 \}\\ \textrm{(or in interval notation:$(-\infty,-4) \cup (-4,1) \cup (1,\infty)$)} \end{array}}$

6. If $\displaystyle{f(x) = \frac{3}{x-1} \textrm{ and } g(x) = 5x+2}$, find the following:

1. The domain of $f$

${\textrm{Noting the denominator must not be zero, the domain is } \{ x \in \mathbb{R}, x \neq 1\}}$

2. The domain of $g$

${\displaystyle{\textrm{all reals, } \mathbb{R}}}$

3. $\displaystyle{(f + g)(x)}$ and it's domain

${ \begin{array}{l} \displaystyle{(f + g)(x) = \frac{3}{x-1} + (5x+2)}\\\\ \textrm{Noting that denominators must not be zero,}\\ \textrm{the domain is } \{ x \in \mathbb{R}, x \neq 1\}\\\\ \textrm{After finding the domain, we can simplify things:}\\ (f + g)(x) = \displaystyle{\frac{3 + (5x+2)(x-1)}{x-1} = \frac{5x^2-3x+1}{x-1}} \end{array}}$

4. $\displaystyle{(f - g)(x)}$ and it's domain

${ \begin{array}{l} \displaystyle{(f - g)(x) = \frac{3}{x-1} - (5x+2)}\\\\ \textrm{Noting that denominators must not be zero,}\\ \textrm{the domain is } \{ x \in \mathbb{R}, x \neq 1\}\\\\ \textrm{After finding the domain, we can simplify things:}\\ (f - g)(x) = \displaystyle{\frac{3 - (5x+2)(x-1)}{x-1} = \frac{-5x^2+3x+5}{x-1}} \end{array}}$

5. $\displaystyle{(f \circ g)(x)}$ and it's domain

${ \begin{array}{l} \displaystyle{(f \circ g)(x) = f(g(x)) = \frac{3}{(5x+2) -1}}\\\\ \textrm{Noting that denominators must not be zero, and solving$(5x+2) - 1 = 0$,}\\ \textrm{we find$x=-1/5$, so the domain is } \{ x \in \mathbb{R}, x \neq -1/5 \}\\\\ \textrm{After finding the domain, we can simplify things:}\\ (f \circ g)(x) = \displaystyle{\frac{3}{5x+1}} \end{array}}$

6. $\displaystyle{\left( \frac{f}{g} \right) (2)}$

${\displaystyle{\left( \frac{f}{g} \right) (2) = \frac{f(2)}{g(2)} = \frac{1}{4}}}$

7. $\displaystyle{(f \cdot g) (x)}$

${\displaystyle{(f \cdot g)(x) = f(x)g(x) = \frac{3(5x+2)}{x-1}}}$

7. If $\displaystyle{f(x) = 2x+2 \textrm{ and } g(x) = \frac{x-3}{x+1}}$, find the following:

1. The domain of $f$

${\displaystyle{\textrm{all reals, } \mathbb{R}}}$

2. $\displaystyle{(f + g)(x)}$ and it's domain

${ \begin{array}{l} \displaystyle{(f + g)(x) = (2x+2) + \frac{x-3}{x+1}}\\\\ \textrm{Noting that denominators must not be zero,}\\ \textrm{the domain is } \{ x \in \mathbb{R}, x \neq -1\}\\\\ \textrm{After finding the domain, we can simplify things:}\\ (f + g)(x) = \displaystyle{\frac{(2x+2)(x+1)+(x-3)}{x+1} = \frac{2x^2+5x-1}{x+1}} \end{array}}$

3. $\displaystyle{(f - g)(x)}$ and it's domain

${ \begin{array}{l} \displaystyle{(f - g)(x) = (2x+2) - \frac{x-3}{x+1}}\\\\ \textrm{Noting that denominators must not be zero,}\\ \textrm{the domain is } \{ x \in \mathbb{R}, x \neq -1\}\\\\ \textrm{After finding the domain, we can simplify things:}\\ (f - g)(x) = \displaystyle{\frac{(2x+2)(x+1)-(x-3)}{x+1} = \frac{2x^2+3x+5}{x+1}} \end{array}}$

4. $\displaystyle{(f \circ g)(x)}$

${ \begin{array}{l} \begin{array}{l} \displaystyle{(f \circ g)(x) = f(g(x)) = 2 \left( \frac{x-3}{x+1} \right) + 2}\\\\ \textrm{Noting that denominators must not be zero,}\\ \textrm{the domain is } \{ x \in \mathbb{R}, x \neq -1\}\\\\ \textrm{After finding the domain, we can simplify things:} \end{array}\\\\ \begin{array}{rcl} (f \circ g)(x) &=& \displaystyle{\frac{2(x-3) + 2(x+1)}{x+1}}\\\\ &=& \displaystyle{\frac{4x-4}{x+1}} \end{array} \end{array}}$

5. $\displaystyle{(f/g)(x)}$ and it's domain

${ \begin{array}{l} \begin{array}{l} \displaystyle{(f/g)(x) = \frac{f(x)}{g(x)} = \frac{2x+2}{\displaystyle{\frac{x-3}{x+1}}}}\\\\ \textrm{Noting that both denominators must not be zero,}\\ \textrm{we solve each of the following equations:}\\\\ \quad \quad x+1 = 0 \quad \textrm{ and } \quad \displaystyle{\frac{x-3}{x+1} = 0}\\\\ \textrm{Finding solutions of$x=-1$and$x=3$, respectively,}\\ \textrm{we arrive at the domain: } \{ x \in \mathbb{R}, x \neq -1,3\}\\\\ \textrm{After finding the domain, we can simplify things:} \end{array}\\\\ \begin{array}{rcl} (f/g)(x) &=& \displaystyle{(2x+2) \cdot \frac{x+1}{x-3}}\\ &=& \displaystyle{\frac{2x^2+4x+2}{x-3}} \end{array} \end{array}}$

6. $\displaystyle{(f - g) (2)}$

${ \begin{array}{rcl} (f-g)(2) &=& f(2) - g(2)\\ &=& \displaystyle{6 + \frac{1}{3}}\\ &=& \displaystyle{\frac{19}{3}} \end{array}}$

8. Find the domain of the function:

${\textrm{Note: } \displaystyle{\{ x \in \mathbb{R} \, | \, x \neq 10,7,\pm \sqrt{2} \textrm{ and } x \in (-5,1)\}} \textrm{ So, more briefly: } (-5,-\sqrt{2}) \cup (-\sqrt{2},1)}$ $$f(x) = \frac{5x+7}{(x^2-2)\sqrt{5-x^2-4x}} + \frac{x-7}{x-7} - \frac{13}{x-10}$$

9. If $f(y) = 2y^2 - 3y$ and $g(t) = t^2 - 9$, solve $(g \circ f)(x) = 0$

${\displaystyle{x = \frac{3 \pm \sqrt{33}}{4}}}$

10. If $f(x) = x(x+2)$ and $\displaystyle{g(x)=\frac{1}{x |x^2 - 3|}}$, find the domain of   $fg$.

${\displaystyle{\{x \in \mathbb{R} \, | \, x \neq 0, \pm \sqrt{3} \}}}$