Let's Learn Additional Practice Related to Functions

Section 5.12 Additional Practice Related to Functions

Exercises Exercises

As directed, determine either the specified function value or determine the solution set to the stated equation.

1.

Determine $k(9)$ where $k(x)=3+4x-x^2\text{.}$

Solution

$k(9)=-42$

2.

Determine $y(-3)$ where $y(t)=5-x-x^3\text{.}$

Solution

$y(-3)=35$

3.

Determine the solution set to $m(x)=5$ where $m(x)=3x-7\text{.}$

Solution

The solution set is $\{4\}\text{.}$

4.

Determine the solution set to $p(t)=-3$ where $p(t)=t^2-7t-47\text{.}$

Solution

The solution set is $\{-4,11\}\text{.}$

5.

Determine $f(11)$ where $f(x)=\frac{x-4}{x+3}\text{.}$

Solution

$f(11)=\frac{1}{2}$

6.

Determine the solution set to $w(t)=6$ where $w(t)=\sqrt{3t-9}\text{.}$

Solution

The solution set is $\{15\}\text{.}$

7.

Determine $w(6)$ where $w(t)=\sqrt{3t-9}\text{.}$

Solution

$w(6)=3$

8.

Determine the solution set to $q(x)=12$ where $q(x)=3+\abs{x-1}\text{.}$

Solution

The solution set is $\{-8,10\}\text{.}$

Work with a function presented in graphical form.

9.

Determine each of the following function values based upon the function $k$ shown in Figure 5.12.1.

$k(2)$
$k(0)$
$k(4)$
$k(-3)$

Determine the solution set to each of the following equations based upon the function $k$ shown in Figure 5.12.1.

$k(x)=1$
$k(x)=4$
$k(x)=5$
$k(x)=6$

$The graph of a function names $k\text{.}$ There are three distinct pieces to the curve, There is a half-line that points leftward ad downward. The half-line terminates at the point $(0,4)\text{.}$ There is an isolated solid dot at the point $(0,-3)\text{.}$ There is a little more than half of a parabola that originates at an open circle located at the point $(2,4)\text{.}$ The top point on the parabolic piece is located at the point $(3,5)\text{.}$ The parabolic piece continues downward from that point with an arrow at the end.$

Figure 5.12.1. $y=k(x)$

Solution

$k(2)$ is undefined.
$k(0)=-3$
$k(4)=4$
$k(-3)=-5$

The solution set is $\{-1,5\}\text{.}$
The solution set is $\{4\}\text{.}$
The solution set is $\{3\}\text{.}$
The solution set is $\emptyset\text{.}$

Determine the domain of each of the following functions. State each domain using interval notation.

10.

$r(x)=4-\sqrt{8-16x}$

Solution

The domain of $r$ is $\left(-\infty,\frac{1}{2}\right]\text{.}$

11.

$w(t)=\sqrt[5]{6t+42}$

Solution

The domain of $w$ is $(-\infty,\infty)\text{.}$

12.

$f(x)=2x^2-7x+8$

Solution

The domain of $f$ is $(-\infty,\infty)\text{.}$

13.

$q(x)=\frac{y-9}{y+10}$

Solution

The domain of $q$ is $(-\infty,-10) \cup (-10,\infty)\text{.}$

14.

$q(t)=\frac{\sqrt{t-12}}{t-3}$

Solution

The domain of $q$ is $[12,\infty)\text{.}$

15.

$y(x)=\frac{x^2-5x+6}{x^2-8x+12}$

Solution

The domain of $y$ is $(-\infty,2) \cup (2,6) \cup (6,\infty)\text{.}$

Determine the domain and range of functions presented in graphical form.

16.

Determine the domain and range of the function $g$ shown in Figure 5.12.2. State the domain and range using interval notation.

$The graph of a function named $g\text{.}$ The function is a little more than the left-half of a downward opening parabola. The vertex is at the point $(-2,5)\text{.}$ The function stops at a point a little higher the $(1,2)\text{.}$ There is an arrow on the left end of the curve.$

Figure 5.12.2. $y=g(x)$

Solution

The domain of $g$ is $(-\infty,1)$and the range is $(-\infty,5]\text{.}$

17.

Determine the domain and range of the function $z$ shown in Figure 5.12.3. State the domain and range using interval notation.

The graph of a function named $z\text{.}$ The function is made up of two line segments. One of the line segments extends from the point $(-5,1)$ to the point $(0,4)$ with an open circle at the left endpoint and a solid dot at the right endpoint. The other line segment extends from the point $(3,2)$ to the point $(6,-1)$ with pen circles at each end point. — Figure 5.12.3. $y=z(x)$

Solution

The domain of $z$ is $(-5,0] \cup (3,6)$and the range is $(-1,4]\text{.}$

Evaluate and simplify each indicated expression.

18.

Determine $f(x+3)$ for the function $f(x)=x^2-x$

Solution

$f(x+3)=x^2+5x+6$

19.

Determine $g(-x)+5$ for the function $g(x)=2x^3-3x^2-6x+2\text{.}$

Solution

$g(-x)+5=-2x^3-3x^2+6x+7$

20.

Determine $3g(4t)$ for the function $\sqrt{t-7}\text{.}$

Solution

$3g(4t)=3\sqrt{4t-7}$

21.

Determine $-2y(t)-7$ for the function $y(t)=-5t^2-11t+3\text{.}$

Solution

$-2y(t)-7=10t^2+22t-13$

22.

Determine $3r(4t-3)$ for the function $r(t)=\frac{3t-5}{t+4}\text{.}$

Solution

$3r(4t-3)=\frac{36t-42}{4t+1}$

23.

Determine $\frac{w(y+5)-w(y-5)}{2}$ for the function $w(y)=4y-7\text{.}$

Solution

$\frac{w(y+5)-w(y-5)}{2}=20$

24.

Evaluate and simplify $(f \circ f)(x)$ where $f(x)=7-x\text{.}$

Solution

$(f \circ f)(x)=x$

25.

Evaluate and simplify $(g \circ f)(x)$ where $f(x)=3x+9$ and $g(x)=5-x^2\text{.}$

Solution

$(g \circ f)(x)=-9x^2-54x-76$

26.

Evaluate and simplify $(f \circ g)(4)$ where $f(x)=3x+9$ and $g(x)=5-x^2\text{.}$

Solution

$(f \circ g)(4)=-24$

27.

Evaluate and simplify $(g \circ g)(8)$ where $g(t)=\frac{3}{t+1}\text{.}$

Solution

$(g \circ g)(8)=\frac{9}{4}$

28.

Evaluate and simplify $(k \circ h)(t)$ where $h(t)=\frac{9t+2}{6}$ and $k(t)=\frac{6t-2}{9}\text{.}$

Solution

$k \circ h)(t)=t$

Determine the difference quotient for each of the following functions. Make sure that you completely simplify each expression.

29.

$f(x)=3x+7$

Solution

The difference quotient is $\frac{f(x+h)-f(x)}{h}=3,\,\,h \neq 0\text{.}$

30.

$g(x)=x^2-5x+1$

Solution

The difference quotient is $\frac{g(x+h)-g(x)}{h}=2x+h-5,\,\, h \neq 0\text{.}$

31.

$k(t)=3-2t^2$

Solution

The difference quotient is $\frac{k(t+h)-k(t)}{h}=-4t-2h,\,\, h \neq 0\text{.}$

32.

$s(t)=-\frac{3}{2t}$

Solution

The difference quotient is $\frac{s(t+h)-s(t)}{h}=\frac{3}{2t^2+2th},\,\, h \neq 0\text{.}$

Determine solution sets to inequalities based upon the graph of a function.

33.

Determine the solution set to $k(x) \geq 1$ based upon the function $k$ shown in Figure 5.12.4. State the solution set using both set-builder notation and interval notation.

Figure 5.12.4. $y=k(x)$

Solution

The solution set is $\{x \mid -1 \leq x \lt 0 \text{ or } 2 \lt x \leq 5\}\text{.}$

The solution set is $[-1,0) \cup (2,5]\text{.}$

34.

Determine the solution set to $k(x) \leq 4$ based upon the function $k$ shown in Figure 5.12.5. State the solution set using both set-builder notation and interval notation.

Figure 5.12.5. $y=k(x)$

Solution

The solution set is $\{x \mid x \leq 0 \text{ or } x \geq 4\}\text{.}$

The solution set is $(-\infty,0] \cup [4,\infty)\text{.}$

35.

Determine the solution set to $k(x) \gt 5$ based upon the function $k$ shown in Figure 5.12.6. State the solution set using both set-builder notation and interval notation.

Figure 5.12.6. $y=k(x)$

Solution

The solution set is $\{\}\text{.}$

The solution set is $\emptyset\text{.}$

36.

Determine the solution set to $k(x) \lt 6$ based upon the function $k$ shown in Figure 5.12.7. State the solution set using both set-builder notation and interval notation.

Figure 5.12.7. $y=k(x)$

Solution

The solution set is $\{x \mid x \leq 0 \text{ or } x \gt 2\}\text{.}$

The solution set is $(-\infty,0] \cup (2,\infty)\text{.}$

Draw the inverse of each graphed function.

37.

The graph of a linear function named $f\text{.}$ The function passes through the points $(0,-1)$ and $(3,0)\text{.}$ — Figure 5.12.8. $\highlightr{y=f(x)}$

Solution

The graph of two linear functions, one named $f$ and the other named $f$-inverse. The function $f$ passes through the points $(0,-1)$ and $(3,0)\text{.}$ The function $f$-inverse passes through the points $(-1,0)$ and $(0,3)\text{.}$ The two lines ar mirror images across the line $y=x$ (which is also graphed). — Figure 5.12.9. $\highlightr{y=f(x)}$ and $\highlight{y=f^{-1}(x)}$

38.

The graph of a function named $g\text{.}$ The function is the right half of an upward opening parabola. A solid dot sits atop the vertex which lies at the point $(-5,-3)\text{.}$ The points $(-3,-2)$ and $(0,3.25)$ also lie on function. — Figure 5.12.10. $\highlightr{y=g(x)}$

Solution

The graph of two function, one named $g$ and the other named $g$inverse. The function $g$ is the right half of an upward opening parabola. The function $g$-inverse s the top half of a rightward opening parabola. A solid dot sits atop the vertex of $g$ which lies at the point $(-5,-3)\text{.}$ A solid dot sits atop the vertex $g$-inverse which lies at the point $(-3,-5)\text{.}$ The points $(-3,-2)$ and $(0,3.25)$ also lie on function $g\text{.}$ The points $(-2,-3)$ and $(3.25,0)$ also lie on function $g$-inverse. The two functions are mirror images across the line $y=x$ (which is also graphed). — Figure 5.12.11. $\highlightr{y=g(x)}$ and $\highlight{y=g^{-1}(x)}$

Given the formula for $f(x)\text{,}$ determine the formula for $f^{-1}(x)\text{.}$

39.

$f(x)=4x-19$

Solution

$f^{-1}(x)=\frac{1}{4}x+\frac{19}{4}$

40.

$f(x)=\frac{\sqrt[3]{2x+8}}{2}-6$

Solution

$f^{-1}(x)=\frac{1}{2}(2x+12)^3-4$

41.

$f(x)=\frac{1-2x}{3x+11}$

Solution

$f^{-1}(x)=\frac{11-x}{3x+2}$

The graph of a function named $f$ is given in Figure 5.12.12. Three functions are stated in terms of $f\text{.}$ Graph each of these functions.

The graph of a function named $f\text{.}$ The function is a line segment extending from the point $(-6,-3)$ to the point $(3,6)\text{.}$ There are solid dots at each end of the line segment. — Figure 5.12.12. $\highlight{y=f(x)}$

42.

$g(x)=\frac{1}{3}f(3x-6)+2$

Solution

The graph of two functions, one named $f$ and the other named $g\text{.}$ The function $f$ is a line segment extending from the point $(-6,-3)$ to the point $(3,6)\text{.}$ There are solid dots at each end of the line segment. The function $g$ is a line segment extending from the point $(0,1)$ to the point $(3,4)\text{.}$ There are solid dots at each end of the line segment. — Figure 5.12.13. $\highlight{y=f(x)}$ and $\highlightr{y=g(x)}$

43.

$h(x)=\frac{2}{3}f\left(\frac{3}{2}(x+1)\right)-1$

Solution

The graph of two functions, one named $f$ and the other named $h\text{.}$ The function $f$ is a line segment extending from the point $(-6,-3)$ to the point $(3,6)\text{.}$ There are solid dots at each end of the line segment. The function $h$ s a line segment extending from the point $(-5,-3)$ to the point $(3,1)\text{.}$ There are solid dots at each end of the line segment. — Figure 5.12.14. $\highlight{y=f(x)}$ and $\highlightr{y=h(x)}$

44.

$k(x)=-f(-(x+2))$

Solution

The graph of two functions, one named $f$ and the other named $k\text{.}$ The function $f$ is a line segment extending from the point $(-6,-3)$ to the point $(3,6)\text{.}$ There are solid dots at each end of the line segment. The function $k$ is a line segment extending from the point $(-5,-6)$ to the point $(4,3)\text{.}$ There are solid dots at each end of the line segment. — Figure 5.12.15. $\highlight{y=f(x)}$ and $\highlightr{y=k(x)}$

One point on a function named $f$ is $(12,-20)\text{.}$ Several functions are defined in terms of $f\text{.}$ For each function, use graphical transformation properties to determine where the stated point on $f$ resides on the graph of the new function.

45.

$g(x)=-f(4(x-3))-8$

Solution

The point $(12,-20)$ for $f$ moves to $(6,12)$ on $g$

46.

$h(x)=\frac{1}{10}f(\frac{1}{5}x+3)$

Solution

The point $(12,-20)$ for $f$ moves to $(45,-2)$ on $h$

47.

$k(x)=2f(-6x-12)+5$

Solution

The point $(12,-20)$ for $f$ moves to $(-4,-35)$ on $k$

48.

$w(x)=-\frac{2}{5}f(\frac{2}{3}(x+8))-3$

Solution

The point $(12,-20)$ for $f$ moves to $(10,5)$ on $w$