Let's Learn Additional Practice Related to Functions

permalink $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.

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10.

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

Solution

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

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11.

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

Solution

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

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12.

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

Solution

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

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13.

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

Solution

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

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14.

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

Solution

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

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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.

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16.

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

permalink $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 $(-\infty,5]\text{.}$

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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. — $y=z(x)$

Solution

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

Evaluate and simplify each indicated expression.

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18.

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

Solution

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

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19.

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

Solution

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

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20.

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

Solution

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

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21.

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

Solution

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

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22.

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

Solution

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

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23.

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

Solution

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

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24.

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

Solution

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

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25.

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

Solution

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

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26.

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

Solution

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

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27.

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

Solution

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

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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.

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29.

$f(x)=3x+7$

Solution

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

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30.

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

Solution

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

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31.

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

Solution

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

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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.

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

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

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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 $\emptyset\text{.}$

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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.

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37.

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

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 $\highlight{y=f^{-1}(x)}$

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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. — $\highlightr{y=g(x)}$

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 $f^{-1}(x)\text{.}$

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39.

$f(x)=4x-19$

Solution

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

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40.

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

Solution

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

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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. — $\highlight{y=f(x)}$

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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 $\highlightr{y=g(x)}$

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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 $\highlightr{y=h(x)}$

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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.

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