Question

1. Find the domain of the given function. (1 point)

Answer 1

1) Find the domain of the given function. 
f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two.

 using a graphical tool     

see the attachment  

the answer is C) x ≥ -3, x ≠ 2  

 2. Identify intervals on which the function is increasing, decreasing, or constant.

g(x) = 2 - (x - 7)2

 using a graphical tool     

see the attachment

 the answer is  C) Increasing: x < 7; decreasing: x > 7

3. Perform the requested operation or operations.

f(x) = 4x + 7, g(x) = 3x2

Find (f + g)(x).

 (f + g)(x) = f(x) + g(x) 

(f + g)(x) = 4x + 7 + 3x^2 

(f + g)(x) = 3x^2 + 4x + 7

 The answer is  C) 4x + 7 + 3x2

 4. Perform the requested operation or operations.

f(x) = x minus five divided by eight. ; g(x) = 8x + 5, find g(f(x)).

 f(x)=(x-5)/8    g(x)=8x+5

g(f(x))=8((x-5)/8)+5=x-5+5=x

the answer is  B) g(f(x)) = x  

 

5. Find f(x) and g(x) so that the function can be described as y = f(g(x)).
y = nine divided by square root of quantity five x plus five. 

 y=f(g(x))=9/((5x+5) ^1/2)

let do

g(x)=5x+5

 

so

f(x)= 9/( x^1/2)

 

the answer is  A) f(x) = nine divided by square root of x. , g(x) = 5x + 5

6. A satellite camera takes a rectangular-shaped picture. The smallest region that can be photographed is a 4-km by 4-km rectangle. As the camera zooms out, the length l and width w of the rectangle increase at a rate of 3 km/sec. How long does it take for the area A to be at least 4 times its original size?

 

Original size- >4km*4km=16 km2

4 times its original size---------------4*(16km2)-----64 Km2----------- > 8 km by 8 Km

 

Therefore

3km----------------------------- 1 sec

(8km-4km)---------------------x

 

X=4/3=1.33 sec

 

The answer is D) 1.33 sec

7. Find the inverse of the function.

f(x) = the cube root of quantity x divided by seven. - 9

 

to solve

replace f(x) with y
switch x and y
solve for y
replace y with f⁻¹(x)

f(x)=((x/7)-9) ^(1/3)

replce f(x) with y

y=((x/7)-9) ^(1/3)

switch x and y

x=((y/7)-9) ^(1/3)

solve for y

x^3=((y/7)-9)

x^3+9=y/7

y=7(x^3+9)

the answer is C) f-1(x) = 7(x3 + 9)  

8. Describe how the graph of y= x2 can be transformed to the graph of the given equation.
y = (x - 14)2 – 9

using a graphical tool     see the attachment  the answer is C) Shift the graph of y = x2 right 14 units and then down 9 units  

9. Describe how to transform the graph of f into the graph of g. 
f(x) = alt='square root of quantity x minus nine.' and g(x) = alt='square root of quantity x plus five. '

f(x)=(x-9) ^1/2  g(x)=(x+5) ^1/2 

using a graphical tool     see the attachment

the answer is C) Shift the graph of f left 14 units

10. If the following is a polynomial function, then state its degree and leading coefficient. If it is not, then state this fact.

f(x) = -16x5 - 7x4 – 6

The answer is B) Degree: 5; leading coefficient: -16

11. Write the quadratic function in vertex form.
y = x2 + 4x + 7 

Complete the square on the right side of the equation

Use the form ax2+bx+cax2+bx+c, to find the values of a, b, and c.

a=1,b=4,c=7

Consider the vertex form of a parabola.

a(x+d)2+e

Find the value of dd using the formula d=b/2a

d=4/(2*1)=2

Find the value of e using the formula e=c−b2/4a

e=7−4=3

Substitute the values of a, d, and e into the vertex form a(x+d)2+e

(x+2)2+3

 The answer is A) y = (x + 2)2+ 3

12. Find the zeros of the function.

f(x) = 3x3 - 12x2 - 15x

using a graphical tool   (see the attachment)

x1=-1

x2=0

x3=5

The answer is C) 0, -1, and 5

13. Find a cubic function with the given zeros.
7, -3, 2

X1=7

X2=-3

X3=2

f(x)=(x-7)(x+3)(x-2)=(x2-4x-21)(x-2)=x3-6x2-13x+42

 

the answer is C) f(x) = x3 - 6x2 - 13x + 42  

14. Find the remainder when f(x) is divided by (x - k).
f(x) = 7x4 + 12x3 + 6x2 - 5x + 16; k = 3 

 f(x)=7(3)4+12(3)3+6(3)2-5(3)+16=946

  The answer is the B) 946  

15. Use the Rational Zeros Theorem to write a list of all potential rational zeros.
f(x) = x3 - 10x2 + 4x - 24

The constant term of

Answer 2

ANSWER TO QUESTION 1

We have the function

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

The domain of this function  refers to all $x$-values for which $f(x)$ is defined.

This is a rational function that is undefined whenever the denominator is equal to zero. As a result of this we need to exclude the x-values that will make the function undefined.

We can find these values by equating the the denominator to zero.

$(x+8)(x-2)\ne 0$

$(x+8)\ne 0\:or\:(x-2)\ne 0$

$(x\ne -8\:or\:x\ne 2 0$

Also this rational function is undefined whenever the expression under the radical sign in the numerator is less than zero.

We must therefore restrict the expression under the square root sign in the numerator to be greater or equal to zero. Thus

$x+3\ge0$

$x\ge-3$

Because of this last restriction, $x\ne -8$ is eliminated from the restrictions.

Therefore the domain of $f(x)=\frac{\sqrt{x+3}} {(x+8)(x-2)}$ is

$x\ge-3$, $x\ne 2$

The correct answer is C

ANSWER TO QUESTION 2

We have the function

$g(x)=2-(x-7)^2$

which is a maximum function written in the vertex form.

It is a maximum function because when we compare to the general vertex form,

$g(x)=a(x-h)^2+k$

$a=-1$

The vertex of the graph is $V(7,2)$.

The x-value of the vertex is $7$

For $x$, the graph is increasing and for $x>\:7$ the graph will be decreasing.

The correct answer is C

ANSWER TO QUESTION 3

$f(x)=4x+7$

and

$g(x)=3x^2$

$(f+g)(x)=f(x)+g(x)$

$(f+g)(x)=4x+7+3x^2$

$(f+g)(x)=4x+7+3x^2$

The correct answer is C

ANSWER TO QUESTION 4

We have

$f(x)=\frac{x-5}{8}$

$g(x)=8x+5$

The composition of the two functions

$g(f(x))$

is given by;

$g(f(x))=g(\frac{x-5}{8})$

This implies that;

$g(f(x))=8\times (\frac{x-5}{8})+5$

$g(f(x))=x-5+5$

$g(f(x))=x$

The correct answer is B.

ANSWER TO QUESTION 5.

Given

$f(g(x))=\frac{9}{\sqrt{5x+5}}$.

Then

$f(x)=\frac{9}{\sqrt{x} }$ and $g(x)=5x+5$

So that;

$f(g(x))=f(5x+5)$

Which will give;

$f(g(x))=\frac{9}{\sqrt{5x+5}}$.

The correct answer is A

ANSWER TO QUESTION 6

The dimension of the smallest region are

$l=4$ km

and

$w=4km$

If the length and width are increasing at the rate of  3km\sec, then after $x$ seconds, the dimensions will be,

$l=4+3x$ km

and

$w=4+3x$km

This means the area after $x$ seconds is given by;

$(4+3x)(4+3x)$.

If the area is at least 4 times its original size, then we can write the inequality;

$(4+3x)(4+3x)\ge 4(4\times 4)$.

$(4+3x)(4+3x)\ge 64$

We expand to obtain;

$9x^2+24x+16ge 64$

We simplify to obtain,

$9x^2+24x-48ge 0$

We split factor to get;

$(x+4)(3x-4)ge 0$

This implies that;

$x\le -4$

or

$x\ge \frac{4}{3}$

Since we are dealing with time, we discard the negative value. Hence  it will   take

$x= \frac{4}{3} \approx 1.33sec$

The correct answer is D.

ANSWER TO QUESTION 8

The correct answer is C.

This is how the following transformations affect the graph of

$y=x^2$

$y=a(x+b)^2+c$

If $a$, the graph opens down.

If $a>\:0$, the graph opens up.

If $b$, the graph graph shifts to the right b units.

If $b>\:0$, the graph graph shifts to the left b units.

If $c$, the graph graph shifts down c units.

If $c>\:0$,the graph graph shifts up c units.

Therefore for $y=(x-14)^2-9$, the graph of $y=x^2$ moves to the right 14 units and moves down 9 units.

ANSWER TO QUESTION 9

The graph of

$f(x)=\sqrt{x-9}$ has y-intercept which is $9$

and

$f(x)=\sqrt{x+5}$ has x-intercept which is $-5$

In order to transform the graph of $f$ in to $g$, we need to shift  the x-intercept of the graph of f 14 units to the left.

The correct  answer is C

ANSWER TO QUESTION 10

The function;

$f(x)=-16x^5-7x^4-6$.

This is a polynomial with a degree of 5 and the leading coefficient is $-16$.

The correct answer is C

ANSWER TO QUESTION 11

We have

$y=x^2+4x+7$

We add and subtract half the coefficient of $x$.

$y=x^2+4x+(2)^2 -(2)^2+7$

The first three term is a perfect square.

$y=(x+2)^2 -4+7$

$y=(x+2)^2 +3$

The correct answer is A.

ANSWER TO QUESTION 12

We want to solve;

$f(x)=3x^3-12x^2-15x$

To find the zeros of this function, we equate it to zero.

$3x^3-12x^2-15x=0$

We factor $3x$ to obtain;

$3x(x^2-4x-5)=0$

We now split the middle term of the quadratic factor to get,

$3x(x^2-5x+x-5)=0$

$3x(x(x-5)+1(x-5)=0$

$3x(x-5)(x+1)=0$

$3x=0,(x-5)=0,(x+1)=0$

$x=0,x=5,x=-1$

The correct answer is C.

See Attachment for the rest of the solutions