Expert: Scott A Wilson - 7/24/2010

Question
QUESTION: Hey, I was doing practice problems and kinda got stuck on a few. Could you please help me, thanks!

Determine whether the function is even, odd, or neither.

1. √7x - 5= 3

Find the zeros of  

2. x^2 - 2x + 15/ 3x
**I'm almost certain there are no zeros for this one.

3. (x^5 – 32)/ x – 4
**I'm almost certain it is x= 2

Describe the transformation

4. f(x) = -(x + 3)^2 - 5

5. f(x) = (x - 2)^2 + 4

6. f(x) = 3/4 (x - 9)^2 + 3

Find the inverse

7. f(x)= x^2 -5

8. f(x) = 2x^5 – 11


Determine the intervals where the graph is increasing, decreasing, and constant. (I've attached the graph)

ANSWER: 1. I see no function, only an equation with a solution for x.
Addding 5 to both sides gives √7x = 2.
If the x is in the squareroot, then squaring both sides gives 7x = 4.
If the x is not in the squareroot, squaring both sides gives 7x² = 4.
Either one can be solved for x; the second is ± when a squareroot is taken.


Find the zeros of  

2. x^2 - 2x + 15/ 3x
**I'm almost certain there are no zeros for this one.
The way to check is take ax² + bx + c = 0 and see what a, b, and c are.
The quadratic equation says there are no real solutions if b²-4ac is negative.

3. (x^5 – 32)/ x – 4
**I'm almost certain it is x= 2
I don't see the question, but 2^5 is 32.

Describe the transformation

4. f(x) = -(x + 3)^2 - 5
Add 3 to the value, square it, take the negative of that, and subtract 5

5. f(x) = (x - 2)^2 + 4
Subtract 2 from the value, square it, and then add 4

6. f(x) = 3/4 (x - 9)^2 + 3
Subtract 9 from the value, square that, multiply by 3/4, and add 3

Find the inverse

7. f(x)= x^2 -5
Take y = x² - 5; the inverse is found by solving for x.
That is, x² = y + 5, so x = ±√(y+5).

8. f(x) = 2x^5 – 11
Here, y = 2x^5 - 11, so 2x^5 = y + 11, so x^5 = (y+11)/2, so x = ((y+11)/2)^(1/5).
Note that 1/5 = 0.2, so this is x = ((y+11)/2)^0.2 or the fifth root of ((y+11)2).

Determine the intervals where the graph is increasing, decreasing, and constant. (I've attached the graph)

The graph is decreasing until x=1 and increasing past x=2.  In between, it is doing neither.


---------- FOLLOW-UP ----------

QUESTION: I'm confused still on a few...the first one, the x is under the square root, but how do I determine if its odd, even, or neither?

Describing the transformation,4-6 I'm not sure either. I don't think thats what the directions were asking for.

Answer
In terms of a fuction of y in terms of x, it is neither since it is not defined for negative x.
It can be looked at as an even function of x in terms of y.

4. f(x) = -(x + 3)^2 - 5
Add 3 to the value, square it, take the negative of that, and subtract 5.
This shifts the function x² to the left and then flips it upside down.

5. f(x) = (x - 2)^2 + 4
Subtract 2 from the value, square it, and then add 4.
This shifts the function x² to the right and then upwards by 4.

6. f(x) = 3/4 (x - 9)^2 + 3
Subtract 9 from the value, square that, multiply by 3/4, and add 3
This shifts the function x² to the right by 9, compacts it horizontally by 3/4,
and then shifts it up by 3.