# Thread: Be merciful and help me.

1. ## Be merciful and help me.

Hello people, next week I have my first ever calculus exam, and I’m so confused, I’m having a hard time trying to understand everything, my professor gave us a review sheet, but I don’t 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 don’t 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 don’t know how to put graphs on here, so I’d say its points, (-3,2) (-1,2)( 0,½) ((3,1)(½,0)(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, what’s f(x)? what point on the graph of F(x)?
-if x=2, what’s 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.

2. 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 don’t know how to put graphs on here, so I’d say its points, (-3,2) (-1,2) ( 0,½) (3,1) (½,0) (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, what’s f(x)? f(4) = -3; what point on the graph of F(x)? Therefore the point is (4, -3)

-if x = 2, what’s 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?

3. Originally Posted by jhonwashington
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,

$f(x)=\frac{x+1}{x-1}$. 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

4. Originally Posted by jhonwashington
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

5. Originally Posted by earboth
Hello,

$f(x)=\frac{x+1}{x-1}$. 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.