9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Textbook Question

Chapter 7, Problem 11CR

Consider the function

$f ( x ) = 2 x 5 − x 4 − 4 x 3 + 2 x 2 + 2 x − 1$

Find the real zeros and their multiplicity.

Find the intercepts.

Find the power function that the graph of $f$ resembles for large $| x | .$

Graph $f$ using a graphing utility.

Approximate the turning points, if any exist.

Use the information obtained in parts $( a ) − ( e )$ to graph $f$ by hand.

Identify the intervals on which $f$ is increasing, decreasing, or constant.

(a)

Expert Solution

To determine

The real zeros of $f(x)=2x5−x4−4x3+2x2+2x−1$ and their multiplicity.

Answer to Problem 11CR

Solution:

The real zeros of $f(x)=2x5−x4−4x3+2x2+2x−1$ are $−1,12,1$ .

$−1$ and $1$ are the zero of the function with multiplicity $2$ whereas $12$ is the zero of the function with multiplicity $1$ .

Explanation of Solution

Given information:

The function, $f(x)=2x5−x4−4x3+2x2+2x−1$

Explanation:

Consider the function $f(x)=2x5−x4−4x3+2x2+2x−1$ .

By rational root theorem,

The divisors of the constant term are $p=±1$ .

The divisors of the leading coefficient are $q=±1, ±2$ .

Then, possible rational zeros of the polynomial are, $pq=±1, ±12$ .

Now, test $x=−1$ using synthetic division.

$−1 2 −1 −4 2 2 −1 −2 3 1 −3 1_ 2 −3 −1 3 −1 0$

Here, since the remainder is $0$ , $x=−1$ is a zero of $f$ .

After taking $x+1$ as a factor,

$2x5−x4−4x3+2x2+2x−1=(x+1)(2x4−3x3−x2+3x−1)$

Then, the depressed equation is $2x4−3x3−x2+3x−1$ .

$2x4−3x3−x2+3x−1$

$=x4+x4−3x3−x2+3x−1$

$=x4−x2+x4−1−3x3+3x$

$=x2(x2−1)+(x2−1)(x2+1)−3x(x2−1)$

$=(x2−1)(x2+x2+1−3x)$

$=(x2−1)(2x2+1−3x)$

$=(x+1)(x−1)(2x2+1−3x)$ .

By quadratic formula, the zeros of the quadratic equation $2x2+1−3x$ ; $x=−b±b2−4ac2a$ .

$x=−(−3)±9−84=3±14=$

$⇒x=1 or x=12$

The factors of $2x2+1−3x$ are $(x−1) and (x−12)$ .

$2x5−x4−4x3+2x2+2x−1=(x+1)(x+1)(x−1)(2x2+1−3x)=(x+1)(x+1)(x−1)(x−1)(x−12)$

The factor form of $f(x)$ is.

$2x5−x4−4x3+2x2+2x−1=(x+1)2(x−1)2(x−12)$

Hence, the real zeros of $f(x)=2x5−x4−4x3+2x2+2x−1$ are $−1,12,1$ and the factor form is $(x+1)2(x−1)2(x−12)$ .

$−1$ is the zero of the function with multiplicity $2$ since the exponent of the factor $(x+1)2$ is $2$ .

$12$ is the zero of the function with multiplicity $1$ since the exponent of the factor $x−12$

is $1$ .

$1$ is the zero of the function with multiplicity $2$ since the exponent of the factor $(x−1)2$ is $2$ .

(b)

Expert Solution

To determine

The $x−$ intercept and $y$ -intercepts of the function, $f(x)=2x5−x4−4x3+2x2+2x−1$ .

Answer to Problem 11CR

Solution:

The $x−$ intercepts of the function are $(0, −1),( 0, 12)$ and, $(0, 1)$ and the $y−$ intercept of the polynomial function is $(−1, 0)$ .

Explanation of Solution

Given information:

The function $f(x)=2x5−x4−4x3+2x2+2x−1$ .

Explanation:

To find $y−$ intercepts substitute $x=0$ in the function $f(x)=2x5−x4−4x3+2x2+2x−1$ ,

$f(0)=2(0)5−(0)4−4(0)3+2(0)2+2(0)−1=−1$

Thus the $y−$ intercept of the polynomial function is $(−1, 0)$ .

Now to find $x−$ intercept of the function substitute $f(x)=0$ in the function.

$f(x)=2x5−x4−4x3+2x2+2x−1$ , it gives

$0=2x5−x4−4x3+2x2+2x−1$

$⇒(x+1)2(x−1)2(x−12)=0$

$⇒ x+1=0$ or, $x−1=0$ or $x−12=0$

$⇒x=−1$ or, $x=1$ or $x=12$

Hence the $x−$ intercepts of the function are $(0, −1),( 0, 12)$ and $(0, 1)$ .

(c)

Expert Solution

To determine

The power function that the graph of $f(x)=2x5−x4−4x3+2x2+2x−1$ resembles for large values of $|x|$ .

Answer to Problem 11CR

Solution:

Thegraph of the function $f(x)=2x5−x4−4x3+2x2+2x−1$ resembles like $y=2x5$ for large values of $|x|$ .

Explanation of Solution

Given information:

The function $f(x)=2x5−x4−4x3+2x2+2x−1$ .

The polynomial function is $f(x)=2x5−x4−4x3+2x2+2x−1$ .

Here the degree of the polynomial function $f(x)$ is $5$ .

The graph of the function $f(x)=2x5−x4−4x3+2x2+2x−1$ behaves like $y=2x5$ for large values of $|x|$ .

(d)

Expert Solution

To determine

To graph: The function $f(x)=2x5−x4−4x3+2x2+2x−1$ using a graphing utility.

Explanation of Solution

Given information:

The function $f(x)=2x5−x4−4x3+2x2+2x−1$ .

Graph:

Use the steps below to graph the function using a graphing calculator.

Step I: Press the ON key.

Step II: Now, press [Y=]. Input the right hand side of the function $y=2x5−x4−4x3+2x2+2x−1$ in $Y1$

Step III: Press [WINDOW] key and set the viewing window as below,

$Xmin= −2Xmax= 2 X scl= 1 Ymin= −2 Ymax= 1 Y scl= 1$ .

Step IV: Then hit [Graph] key to view the graph.

The graph of the function is as follows:

Precalculus, Chapter 7, Problem 11CR , additional homework tip 1

Interpretation:

The graph of the function $f(x)$ crosses the $x−$ axis at $x=12$ since the multiplicity of the $12$ is $1$ that is odd multiplicity.

(e)

Expert Solution

To determine

The approximation of the turning points, if exists, of the function $f(x)=2x5−x4−4x3+2x2+2x−1$ .

Answer to Problem 11CR

Solution:

The turning points of $f(x)=2x5−x4−4x3+2x2+2x−1$ are $(−0.29, −1.325),(1,0)$ , $(0.69, 0.104),(−1,0)$ .

Explanation of Solution

Given information:

The function $f(x)=2x5−x4−4x3+2x2+2x−1$ .

Explanation:

Let, the function $f(x)=2x5−x4−4x3+2x2+2x−1$ .

The maximum number of real zeros is the degree of the polynomial.

Here, the degree of $f(x)=2x5−x4−4x3+2x2+2x−1$ is $5$ .

Since the polynomial function $f(x)=2x5−x4−4x3+2x2+2x−1$ has degree $5$ so the maximum number of the turning points on the graph of the function $f(x)$ is $5−1=4$ .

For the approximation of the turning points find out the maxima and minima using a graphing calculator.

To graph the function $f(x)=2x5−x4−4x3+2x2+2x−1$ using graphing utility use the below steps.

Step I: Press the ON key.

Step II: Now, press [Y=]. Input the right hand side of the function $y=2x5−x4−4x3+2x2+2x−1$ in $Y1$

Step III: Press [WINDOW] key and set the viewing window as below,

$Xmin= −2Xmax= 2 X scl= 1 Ymin= −2 Ymax= 1 Y scl= 1$ .

Step IV: Then hit [Graph] key to view the graph.

The graph of the function is as follows:

Precalculus, Chapter 7, Problem 11CR , additional homework tip 2

To find local maximum and local minimum on the graph using graphing utility use below steps,

Step IV: Press [2ND] [TRACE] to access the calculate menu

Step V: press [MAXIMUM] and press [ENTER].

Step VI: Set left bound by using the left and right arrow. Click [ENTER].

Step VII: Set right bound by using the left and right arrow. Click [ENTER].

Step VIII: Click [Enter] button twice.

It will give the maximum value $x=0.69,y=0.104$

It will give the maximum value $x=−1,y=0$

Thus, the function have its local maximum value at $(0.69, 0.104),(−1,0)$ .

To find local minimum value use below steps.

Step IX: Press [2ND] [TRACE] to access the calculate menu

Step X: press [MINIMUM] and press [ENTER].

Step XI: Set left bound by using the left and right arrow. Click [ENTER].

Step XII: Set right bound by using the left and right arrow. Click [ENTER].

Step XIII: Click [Enter] button twice.

It will give the minimum value $x=−0.29,y=−1.325$ .

It will give the minimum value $x=1,y=0$ .

Thus, the function has its local minimum value at $(−0.29, −1.325),(1,0)$ .

Therefore, the turning points are $(−0.29, −1.325),(1,0)$ , $(0.69, 0.104),(−1,0)$ .

(f)

Expert Solution

To determine

To graph: The function $f(x)=2x5−x4−4x3+2x2+2x−1$ .

Explanation of Solution

Given information:

The function $f(x)=2x5−x4−4x3+2x2+2x−1$

Graph:

The polynomial function is $f(x)=2x5−x4−4x3+2x2+2x−1$ .

From all the above parts, the analysis of the function $f(x)=2x5−x4−4x3+2x2+2x−1$ are stated below:

The graph of the function $f(x)=2x5−x4−4x3+2x2+2x−1$ behaves like $y=2x5$ for large values of $|x|$ .

Thezeros of the function are $−5,−12$ and $3$

The $x−$ intercepts of the function are $−1,12$ and $1$ and the $y−$ intercept of the polynomial function is $−1$ .

The graph of the function $f(x)$ crosses the $x−$ axis at $x=12$ since the multiplicity of the $12$ is $1$ that is odd multiplicity and also the graph of function $f(x)$ touches the $x−$ axis at $x=1,x=−1$ since the multiplicity of the $−1, 1$ are $2$ which is even.

Here the degree of the polynomial function $f(x)$ is $5$ , the maximum number of tuning points are $5−1=4$ which are at $x=−1,x=0.29,x=0.69,x=1$ .

Using all this information, the graph will look alike:

Precalculus, Chapter 7, Problem 11CR , additional homework tip 3

Now find additional points on the graph on each side of $x−$ intercept as follows

For $x=−1.1$ the value of $f(x)$ at $x=−1.1$ is $f(−1.1)=2(−1.1)5−(−1.1)4−4(−1.1)3+2(−1.1)2+2(−1.1)−1=−0.1411$

For $x=−0.5$ the value of $f(x)$ at $x=−0.5$ is $f(−0.5)=2(−0.5)5−(−0.5)4−4(−0.5)3+2(−0.5)2+2(−0.5)−1=−1.125$

For $x=−0.25$ the value of $f(x)$ at $x=−0.25$ is $f(−0.25)=2(−0.25)5−(−0.25)4−4(−0.25)3+2(−0.25)2+2(−0.25)−1=−1.3183$

For $x=0.25$ the value of $f(x)$ at $x=0.25$ is $f(0.25)=2(0.25)5−(0.25)4−4(0.25)3+2(0.25)2+2(0.25)−1=−0.4395$

For $x=0.55$ the value of $f(x)$ at $x=0.55$ is $f(0.55)=2(0.55)5−(0.55)4−4(0.55)3+2(0.55)2+2(0.55)−1=0.0487$

For $x=0.75$ the value of $f(x)$ at $x=0.75$ is $f(0.25)=2(0.25)5−(0.75)4−4(0.75)3+2(0.75)2+2(0.75)−1=0.0957$

For $x=1.1$ the value of $f(x)$ at $x=1.1$ is $f(1.1)=2(1.1)5−(1.1)4−4(1.1)3+2(1.1)2+2(1.1)−1=0.0529$

Now plot all these coordinates $(−1.1, −0.1411), (−0.5, 1.125), (−0.25, −1.3183), (0.25, −0.4395),(0.55,0.0487),(0.75,0.0957),(1.1,0.0529)$ on the graph and join them.

Therefore, the graph of the function is as follows:

Precalculus, Chapter 7, Problem 11CR , additional homework tip 4

Interpretation:

The graph of the function $f(x)=2x5−x4−4x3+2x2+2x−1$ behaves like $y=2x5$ for large values of $|x|$ .

Thezeros of the function are $−5,−12$ and $3$ .

The $x−$ intercepts of the function are $−1,12$ and, $1$ and the $y−$ intercept of the polynomial function is $−1$ .

Here the degree of the polynomial function $f(x)$ is $5$ , the maximum number of tuning points are $5−1=4$ which are at $x=−1,x=0.29,x=0.69,x=1$ .

(g)

Expert Solution

To determine

The intervals where the function $f(x)=2x5−x4−4x3+2x2+2x−1$ is increasing, or decreasing or constant.

Answer to Problem 11CR

Solution:

The function $f(x)=2x5−x4−4x3+2x2+2x−1$ is increasing in the interval $(−∞,−1),(−0.29,0.69),(1,∞)$ and decreasing in the intervals $(−1,−0.29),(0.69,1)$ and nowhere constant.

Explanation of Solution

Given information:

The function, $f(x)=2x5−x4−4x3+2x2+2x−1$ .

The polynomial function is $f(x)=2x5−x4−4x3+2x2+2x−1$ .

From parts (d),(e),(f) the graph of the function $f(x)=2x5−x4−4x3+2x2+2x−1$ is stated below:

Precalculus, Chapter 7, Problem 11CR , additional homework tip 5

Here the degree of the polynomial function $f(x)$ is $5$ , the maximum number of tuning points are $5−1=4$ which are at $x=−1,x=0.29,x=0.69,x=1$ .

From the graph, it is clearly evident the graph is increasing in the interval $(−∞,−1),(−0.29,0.69),(1,∞)$ and decreasing in the intervals $(−1,−0.29),(0.69,1)$ and nowhere constant.

Chapter 7, Problem 10CR

Chapter 7, Problem 12CR

Chapter 7 Solutions

Precalculus

Ch. 7.1 - Prob. 1AYU Ch. 7.1 - Prob. 2AYU Ch. 7.1 - Prob. 3AYU Ch. 7.1 - Prob. 4AYU Ch. 7.1 - Prob. 5AYU Ch. 7.1 - Prob. 6AYU Ch. 7.1 - y= sin 1 x means _____, where 1x1 and 2 y 2 .Ch. 7.1 - Prob. 8AYU Ch. 7.1 - Prob. 9AYU Ch. 7.1 - Prob. 10AYU

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