# Thread: Ap pre-calculus .... Help!!!!

1. ## Ap pre-calculus .... Help!!!!

1. domain and range of

A. y = (-4)/(square root of (x + 1)

and

B. y = square root of (9 + x)

2. A cone is constructed from a circular piece of paper with a 3-inch radius by cutting out a sector of the circle with arc length x. The two edges of the remaining portion are joined together to form a cone with radius r and height h, as shown in the figure. Express the volume V of the cone as a function of x.

3. Determine if the function has an inverse function.

A. y = X^2 + 5x - 3

and

B. y = x^3 - 3x + 5

4. Graph

y = 2 sin (4x + (pie/3)) + 1

2. Originally Posted by psychfan29
1. domain and range of

A. y = (-4)/(square root of (x + 1)

and

B. y = square root of (9 + x)
The domain is the set of values of x that the expression is valid for. Can you think of any restrictions on $\displaystyle \sqrt{x + 1}$ might have on x? (Can the number inside the square root be negative?) Also note that the denominator of an expression can't be 0....

-Dan

3. Originally Posted by psychfan29
3. Determine if the function has an inverse function.

A. y = X^2 + 5x - 3

and

B. y = x^3 - 3x + 5
It can only be a function if it passes the vertical line test. It can only be a 1 to 1 function (an inverse function exists) if it passes the horizontal line test.

-Dan

4. Originally Posted by psychfan29
4. Graph

y = 2 sin (4x + (pie/3)) + 1
You haven't learned how to spell $\displaystyle \pi$ yet??? I'll give you a hint: You don't eat it!

$\displaystyle y = A~sin(kx + \phi) + B$
has an amplitude A, a vertical shift B, a wave number k, and a phase angle $\displaystyle \phi$. What this all means is that
The function is centered about the line y = B
The function has a vertical span of A above y = B and A below y = B
The function has a wavelength (distance between maxima) of $\displaystyle \lambda = 2 \pi / k$.
The function has been displaced $\displaystyle \phi / k$ units to the left.

-Dan