Uhhh I think it'd be C and... probably A or D. I'm not 100% sure, but if I was taking the test, that is what I would choose.
Expand the following:
(x - 6) (3 x^2 + 10 x - 1)
Hint: | Multiply out (x - 6) (3 x^2 + 10 x - 1).
| | | | x | - | 6
| | 3 x^2 | + | 10 x | - | 1
| | | | -x | + | 6
| | 10 x^2 | - | 60 x | + | 0
3 x^3 | - | 18 x^2 | + | 0 x | + | 0
3 x^3 | - | 8 x^2 | - | 61 x | + | 6:
Answer: 3 x^3 - 8 x^2 - 61 x + 6 Thus B:
<u>SOLUTION:</u>
General Equation: y = mx + b
.
Substitute slope = -2 into the equation:
y = -2x + b
.
Find the y-intercept:
4 = -2(-3) + b
4 = 6 + b
b = -2
.
Substitute b = -2 into the equation:
y = -2x - 2
.
Answer: y = -2x - 2
You do the implcit differentation, then solve for y' and check where this is defined.
In your case: Differentiate implicitly: 2xy + x²y' - y² - x*2yy' = 0
Solve for y': y'(x²-2xy) +2xy - y² = 0
y' = (2xy-y²) / (x²-2xy)
Check where defined: y' is not defined if the denominator becomes zero, i.e.
x² - 2xy = 0 x(x - 2y) = 0
This has formal solutions x=0 and y=x/2. Now we check whether these values are possible for the initially given definition of y:
0^2*y - 0*y^2 =? 4 0 =? 4
This is impossible, hence the function is not defined for 0, and we can disregard this.
x^2*(x/2) - x(x/2)^2 =? 4 x^3/2 - x^3/4 = 4 x^3/4 = 4 x^3=16 x^3 = 16 x = cubicroot(16)
This is a possible value for y, so we have a point where y is defined, but not y'.
The solution to all of it is hence D - { cubicroot(16) }, where D is the domain of y (which nobody has asked for in this example :-).
(Actually, the check whether 0 is in D is superfluous: If you write as solution D - { 0, cubicroot(16) }, this is also correct - only it so happens that 0 is not in D, so the set difference cannot take it out of there ...).
If someone asks for that D, you have to solve the definition for y and find that domain - I don't know of any [general] way to find the domain without solving for the explicit function).
Step-by-step explanation:
