a f(a) is the function f(x) where x is replaced by a.
So we have
(3x^2 + x + 3 - (3a^2 + a + 3)) / ( x - a)
= (3x^2 - 3a^2 + x - a) / (x = a) Answer
b (3(x + h)^2 + x + h + 3 - (3x^2 + x + 3)) / h
= (3x^2 + 6xh + 3h^2 + x + h - 3x^2 - x - 3) ) / h
= (6xh + h + 3 h^2) / h
= 6x + 3h + 1 Answer
Answer:
7 hours
Step-by-step explanation:
56/8=7
Answer:
Below.
Step-by-step explanation:
Let the number of bananas I ate be x.
Number of apples = 2x.
Number of blueberries = 3x.
So we have the equation:
x + 2x +3x = 12
6x = 12
x = 2.
So I ate 2 bananas, 4 apples and 6 blueberries.
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).
