College Alegbra 6

ASSIGNMENT 6

MA240 College Algebra

Directions: Be sure to make an electronic copy of your answer before submitting it to Ashworth College for grading. Unless otherwise stated, answer in complete sentences, and be sure to use correct English spelling and grammar. Sources must be cited in APA format. Your response should be a minimum of one (1) single-spaced page to a maximum of two (2) pages in length; refer to the "Assignment Format" page for specific format requirements.

NOTE: Show your work on the problems.

1. Determine whether the equation defines a function with independent variable x. If it does, find the domain. If it does not, find a value of x to which there corresponds more than one value of y.

x|y| = x + 9

Answer

The equation x|y| = x + 9 defines a function with independent variable x,

Below is the explanation for the same.

When we simplify equation x|y| = x + 9

It becomes |y| = 1+9/x

This equation defines a function with independent variable x

To find its domain we need to look for values of x for which 1+9/x is always a positive real number.

So for any value of x which is less than -9, 1+9/x will always be a positive real number.

And similarly for all real values of x which are greater than zero, 1+9/x will always be a positive real number.

So domain of |y| = 1+9/x is -∞ > x < -9 and 0 < x < ∞.

2. Use the graph of the function to estimate:

a. f(2)

b. f(–4)

c. All x such that f(x) = 0

Answer :

By observing the graph we can find value of f(x) at different points

A. In order to determine value of f(x) at x = 2 or f(2) ,

If we observe the graph then value of f(x) corresponding to x = 2 is 6

Since at x = 2, f(x) cuts the y = 6 line.

B. In order to determine value of f(x) at x = -4 or f(-4) ,

If we observe the graph then value of f(x) corresponding to x = -4 is 0.

Since at x = -4, f(x) cuts the y = 0 line.

C. In order to determine all values of x such that f(x) = 0,

If we observe the graph then value of f(x) becomes zero only when x = -4

So, for x = -4 f(x) = 0.

3. For the following graph:

a. Find the domain of f.

b. Find the range of f.

c. Find the x-intercepts.

d. Find the y-intercept.

e. Find the intervals over which f is increasing.

f. Find the intervals over which f is decreasing.

g. Find the intervals over which f is constant.

h. Find any points of discontinuity.

Answer:

by observing the graph we find

A. Domain of f is all real x or -∞ < x < ∞

B. range of 'f’ is -∞ < f <= 4,

Since its value varies between all real values <=4.

C. x-intercepts are -1, 3,

Since the graph cuts x axis at x = -1 and x = 3.

D. y-intercept is 3.

Since the graph cuts y axis at y = 3.

E. by observing the graph we can determine that,

f is increasing for -∞ < x < 4

F. by observing the graph we can determine that,

f is decreasing for 4 < x < ∞

G.graph of f is either increasing or decreasing, its never constant.

H. by observing the graph we can determine that,

graph has no point of discontinuity

4. Use the following to answer questions a-d:

f(x) = x2 + x and g(x) = x – 5

a. Find h(x) = (f ◦ g)(x).

b. State the domain of h(x) = (f ◦ g)(x).

c. Find h(x) = (g ◦ f)(x).

d. State the domain of h(x) = (g ◦ f)(x).

Answer:

A. h(x) = (f*g)(x)

· h(x) = f(g(x)) by definition of composition

· h(x) = (g(x))2+g(x)

· h(x) = (x-5)2+(x-5) = x2-10x+25+x-5 = x2-9x+20

So h(x) = x2-9x+20

B. Since value of b2-4ac > 0 for h(x)

Hence domain of h(x) is All real x or -∞ < x < ∞

C. h(x) = (g*f)(x)

· h(x) = g(f(x))

· h(x) = f(x) – 5

· h(x) = x2+x-5

D. Since value of b2-4ac > 0 for h(x)

Hence domain of h(x) is All real x or -∞ < x < ∞

Below are the problems you missed, the amount of points they're worth, and the reference page the problem is located on: 1 - 10pts - 236 3e,f - 6pts each - 250-260