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Math Help - Be merciful and help me.

  1. #1
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    Be merciful and help me.

    Hello people, next week I have my first ever calculus exam, and Iím so confused, Iím having a hard time trying to understand everything, my professor gave us a review sheet, but I donít know how to solve those exercises, could somebody please be that nice to give me a hand and explain me easily how to solve them.

    This is the review sheet


    1-For the given function f defined by the equation f(x) =3x^2+2x-4, find

    A f(x+1)
    B f (-x) c -f(x)

    2-Given f(x) = -3x+1, find: F(x+h) - f(x)/h ( in case you donít understand, / is divided by h)

    3-Find the domain of each function
    a) f(x) = x/x^2-3x-4
    b- g(x) = x-2/x^3+x
    c- H(x) √ 2 -6x

    4- Given f(x) = x+1/x-1 and g(x) = 1/x, find the following and state the domain of the functions, be sure to simplify

    A (f-g)(x)
    B (f/g) (x)

    5, in this point they give me a graph, I donít know how to put graphs on here, so Iíd say its points, (-3,2) (-1,2)( 0,Ĺ) ((3,1)(Ĺ,0)(1,-1) and (2,-1) and having this graph as model they ask me, to find
    a) its domain and range
    B- the open intervals on which it is increasing, decreasing, or constant
    C- The intercepts, if any

    6 Determine (algebraically) Wheter the given function is even, odd or neither F(x) = 5x/1+x^2

    7Find the average rate of change from 2 to x, for the given function f, be sure to simplify, F(x) = 3x-4x^2
    8they give me the function f(x) = x+2/x-6 and ask me
    -Is the point (3,14) on the graph of f?
    -if x=4, whatís f(x)? what point on the graph of F(x)?
    -if x=2, whatís f(x)? what point on the graph of F(x)?
    -What is the domain of F?
    - list the x-intercepts, if any, of the graph of F.
    _list the y intercept, if there is one, of the graph of f


    8, find the function that is finally graphed after the following transformation are applied to the graph y =x^2
    - Shifted upward by a factor of 4 and to the right by a factor of 2
    - Reflected about the x axis
    = Shift down 3 units

    9 Graph the given function using the techniques of shifting. Label at least three points

    a-F(x) = |x+1| -2
    b- F(x) =

    10 the price p and the quantity x sold of a certain product obet the demand equation x = -20p+500, 0 >= p >=25

    A Express the revenue R as the function of x
    B what is the revenue if 20 units are sold?


    Thank you so much.
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  2. #2
    Junior Member AlvinCY's Avatar
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    Whoa. Ok... I can only help you with so much of this:


    1- For the given function f defined by the equation f(x) = 3x^2 + 2x - 4, find

    A) f(x+1) = 3(x + 1)^2 + 2(x + 1) - 4 = 3x^2 + 8x + 1

    B) I don't understand your notation, what's the c? I'm assuming it's a typo for an x (multiplication)

    f(-x) = 3(-x)^2 + 2(-x) - 4 = 3x^2 - 2x - 4
    f(-x) \times -f(x) = (3x^2 + 2x - 4)(-3x^2 - 2x + 4) Expand brackets and group like terms, I trust that you know how to do that. Otherwise, you're in big trouble for this exam.

    2- Given f(x) = -3x+1, find: \frac{f(x + h) - f(x)}{h}

    f(x+h) = -3(x+h) + 1

    So \frac{f(x + h) - f(x)}{h}
    = \frac{-3(x + h) + 1 - (-3x + 1)}{h}
    = \frac{-3h}{h}
    = -3

    3- Find the domain of each function

    a) f(x) = \frac{x}{x^2 - 3x - 4}

    The only restriction is that x^2-3x-4 \not = 0
    That is: (x - 4)(x + 1) \not = 0,  x \not = 4, x \not = -1 is the domain restriction

    b) g(x) = \frac{x - 2}{x^3 + x}

    Once again the only restriction is that x^3 + x \not= 0
    That is:  x(x^2 + 1) \not = 0, x \not = 0 as (x^2 + 1) \not = 0 anyway

    c) H(x) = \sqrt {2 - 6x}

    The only restriction is you can square root a negative number so 2 - 6x \ge 0, therefore x < \frac{1}{3}

    4- Given f(x) = x+1/x-1 and g(x) = 1/x, find the following and state the domain of the functions, be sure to simplify

    A (f-g)(x) - Sorry... I don't think I've ever seen notation like that...
    B (f/g) (x) - Sorry... I don't think I've ever seen notation like that...

    5- In this point they give me a graph, I donít know how to put graphs on here, so Iíd say its points, (-3,2) (-1,2) ( 0,Ĺ) (3,1) (Ĺ,0) (1,-1) and (2,-1) and having this graph as model they ask me, to find

    The points make no sense unless you put it up as a graph

    6- Determine (algebraically) Wheter the given function is even, odd or neither F(x) = \frac{5x}{1 + x^2}

    For a function to be odd F(x) = -F(-x), for a function to be even, F(x) = F(-x). And if it's neither cases, then it's neither.

    F(-x) = \frac{-5x}{1 + (-x)^2}
    = -\frac{5x}{1 + x^2} = -F(x), therefore this function is odd.

    7- Find the average rate of change from 2 to x, for the given function f, be sure to simplify, f(x) = 3x - 4x^2

    At x = 2, f(x) = -10

    My best guess at this question (cos I have no idea what the question is asking, unless they're asking for an integral?) would be...

    \frac{3x - 4x^2 + 10}{x - 2} =  - 4x - 5

    8- They give me the function [tex]f(x) = \frac{x + 2}{x - 6} and ask me

    -Is the point (3,14) on the graph of f?

    Is 14 = \frac{3 + 2}{3 - 6}? And the answer is no, so no, (3, 14) doesn't lie on the graph f.

    -if x = 4, whatís f(x)? f(4) = -3; what point on the graph of F(x)? Therefore the point is (4, -3)

    -if x = 2, whatís f(x)? f(2) = -1 what point on the graph of F(x)? Therefore the point is (2, -1)

    -What is the domain of F?

    Domaing is restricted by x - 6 \not = 0 (denominator of a fraction cannot be 0, therefore  x \not = 6

    - List the x-intercepts, if any, of the graph of F.

    x-intercept occurs when f(x) = 0 and it occurs when x = -2, so (-2, 0) is the x-intercept

    - List the y intercept, if there is one, of the graph of f

    y-intercept occurs when x = 0, and gives f(x) = -1/3, therefore (0, -1/3) is the y-intercept


    9- Find the function that is finally graphed after the following transformation are applied to the graph y = x^2

    - Shifted upward by a factor of 4 and to the right by a factor of 2

    y = (x-2)^2 + 4

    To shift a graph f(x) up by a factors, it's f(x) + a, likewise with downwards, it's f(x) - a, to shift the graph left a factors, its f(x + a), and to shift to the right, it's f(x -a).

    - Reflected about the x axis

    Replacing f(x) with -f(x)... so y = x^2 when reflected about the x-axis will become y = -x^2

    - Shift down 3 units

    y = x^2 - 3

    10- Graph the given function using the techniques of shifting. Label at least three points

    a) F(x) = |x+1| - 2

    Graph G(x) = |x| which is just a V shape, origin at (0, 0), then shift this to the left a space and own two spaces (so move the origina to (-1, -2))

    b) You didn't type a function...

    11- the price p and the quantity x sold of a certain product obey the demand equation x = -20p + 500, 0 \ge p \ge 25

    I haven't touched on Revenues and stuff in High School so I'm afraid I can't help you here, sorry. I could only guess a way of doing it... So I'll leave it for someone who knows.

    A) Express the revenue R as the function of x
    B) what is the revenue if 20 units are sold?
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  3. #3
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    Quote Originally Posted by jhonwashington View Post
    Hello people, ...could somebody please be that nice to give me a hand and explain me easily how to solve them.
    This is the review sheet
    ...
    4- Given f(x) = x+1/x-1 and g(x) = 1/x, find the following and state the domain of the functions, be sure to simplify

    A (f-g)(x)
    B (f/g) (x)

    ...
    Hello,

    I assume that your function f reads:

    f(x)=\frac{x+1}{x-1}. If so then

    (f-g)(x)=\frac{x+1}{x-1}-\frac{1}{x}=\frac{x(x+1)}{x(x-1)}-\frac{1 \cdot (x-1)}{x(x-1)}=  \frac{x(x+1)-(x-1)}{x(x-1)}=\frac{x^2+x-x+1}{x(x-1)}=\frac{x^2+1}{x(x-1)}


    \left(\frac{f}{g} \right)(x)=\frac{\frac{x+1}{x-1}}{\frac{1}{x}}=\frac{x(x+1)}{1 \cdot (x-1)}=\frac{x^2+x}{x-1}


    EB
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  4. #4
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    Quote Originally Posted by jhonwashington View Post
    Hello people... could somebody please be that nice to give me a hand and explain me easily how to solve them.
    ...
    10 the price p and the quantity x sold of a certain product obet the demand equation x = -20p+500, 0 >= p >=25

    A Express the revenue R as the function of x
    B what is the revenue if 20 units are sold?
    Hello,

    first express the price p by x:

    x = -20p+500 \Longleftrightarrow p=25-\frac{1}{20}x

    The revenue = quantity * price. Thus:

    R(x)= p \cdot x = \left( 25-\frac{1}{20}x \right) \cdot x= 25x-\frac{1}{20}x^2

    To calculate the revenue at a quantity of 20 you have two possible ways to do:

    a) x = 20 therefore p(20)=25-\frac{1}{20} \cdot 20 = 24. Thus R = 20 * 24 = 480

    b) calculate R(20): R(20)=  25 \cdot 20-\frac{1}{20}\cdot 20^2=500-20=480

    EB
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  5. #5
    Junior Member AlvinCY's Avatar
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    Quote Originally Posted by earboth View Post
    Hello,

    I assume that your function f reads:

    f(x)=\frac{x+1}{x-1}. If so then

    (f-g)(x)=\frac{x+1}{x-1}-\frac{1}{x}=\frac{x(x+1)}{x(x-1)}-\frac{1 \cdot (x-1)}{x(x-1)}=  \frac{x(x+1)-(x-1)}{x(x-1)}=\frac{x^2+x-x+1}{x(x-1)}=\frac{x^2+1}{x(x-1)}


    \left(\frac{f}{g} \right)(x)=\frac{\frac{x+1}{x-1}}{\frac{1}{x}}=\frac{x(x+1)}{1 \cdot (x-1)}=\frac{x^2+x}{x-1}


    EB

    Ah... I figured that's what it's asking, you see, in Sydney we use the notation f(x) - g(x) and f(x)/g(x)... we don't put the f's and g's together.
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