Calculus Review Problems

[lalalina]

This is part of a calculus readiness review. No calculuators are allowed.

1. Given f(x) = sqrt(9-9x) ,
a) Find the domain of f (all real values of x such that f(x) is real) and find the range of f.
The domain is (-inf, 1].
I don't know how to solve for range. Is it just [0, +inf) because it's a square root function?

b) Solve for x in terms of y in the equation y = sqrt(9-9x).
y^2 = 9 - 9x
9x = 9 - y^2
x = (9 - y^2)/9
x = 1 - (y^2/9)
Is this correct?

2. Solve for x: (1-x)^-1/2 - (1-x)^1/2 = 0.
I don't know how to solve for this? I know you're supposed to multiply by something like sqrt(1-x) on top or on bottom...

3. Is cos(5 pi / 6) = (sqrt3)/2?

4. Solve for x: 3x = 1 - 2x^2.
I get 2x^2 + 3x - 1= 0. Now what? It does not factor...

THANK YOU FOR ALL REPLIES!

 

[mmm4444bot]

1. Given f(x) = sqrt(9-9x) ,
a) Find the ... range of f.

Is it just [0, +inf) because it's a square root function?

Hi Lalalina:

Yes, in part.

[0, ?) is the range of f, but not for the reason you state.

Counterexample: g(x) = sqrt(R) can also be interpreted as a "square root function", but zero is NOT in the range of g when R is 9/x. In other words, not all functions defined as

f(x) = sqrt(something)

have a range of [0, ?). So, be careful.

~ Mark

 

[mmm4444bot]

PART B OF EXERCISE NUMBER 1

b) Solve for x in terms of y ... y = sqrt(9-9x).

x = 1 - (y^2/9)
Is this correct?

Yup, it sure is! Good job.

~ Mark

 

[mmm4444bot]

EXERCISE NUMBER 2

2. Solve for x: (1-x)^-1/2 - (1-x)^1/2 = 0.
I don't know how to solve for this ...

I think that you do know how to solve this.

You know how to rewrite the following expression as a single ratio, right?

\(\displaystyle \frac{1}{a} - \frac{a}{1}\)

Please think about execise number 2 some more ...

\(\displaystyle \frac{1}{\sqrt{1 - x}}\;-\;\frac{\sqrt{1 - x}}{1}\;=\;0\)

~ Mark

 

[mmm4444bot]

EXERCISE NUMBER 3

3. Is cos(5 pi / 6) = (sqrt3)/2?

You forgot to type your question about this exercise.

There is A LOT OF TRIGONOMETRY to learn before calculus. Have you taken a basic course in trigonometry? I do not have any idea what you know about the cosine function.

When I read this exercise, my mind jumped through the following sequence ...

• 5 * Pi/6 RADIANS is 150°[/*:m:2o21bnx4]
• The angle 150° drawn in standard position falls into the second quadrant[/*:m:2o21bnx4]
• The sign of cos(150°) must therefore be negative[/*:m:2o21bnx4]
• ?3/2 is not negative ...[/*:m:2o21bnx4]

There are several other ways to approach review question 3.

 

[mmm4444bot]

EXERCISE NUMBER 4

4. Solve for x: 3x = 1 - 2x^2.
I get [that the given equation is equivalent to] 2x^2 + 3x - 1= 0 ...

You're 100% accurate! Good start.

... Now what? It does not factor ...

Use a solution method that does not require factoring the polynomial; there are different methods from which to choose. Have you heard of "completing the square"?

These four review exercises are very BASIC for a calculus student. I'm concerned that you may struggle with calculus, NOT because the concepts in introductory calculus are difficult, BUT because it appears to me that you will be spending most of your mental resources during this class trying to learn basic algebra and trigonometry at the same time.

Anyway, here's my work for completing the square ...

~ Mark