Describe how to find the average rate of change between x=3 and x=5 for the function f(x)= 3x^3+2
1 answer:
You might be interested in
Answer:
The equation determine a relation between x and y
x = ±
y = ±
The domain is 1 ≤ y ≤ 3
The domain is -1 ≤ x ≤ 1
The graphs of these two function are half circle with center (0 , 2)
All of the points on the circle that have distance 1 from point (0 , 2)
Step-by-step explanation:
* Lets explain how to solve the problem
- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where
R is read as "has distance 1 of"
- This relation can also be read as “the point (x, y) is on the circle
of radius 1 with center (0, 2)”
- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”
* <em>Lets solve the problem</em>
- The equation of a circle of center (h , k) and radius r is
(x - h)² + (y - k)² = r²
∵ The center of the circle is (0 , 2)
∴ h = 0 and k = 2
∵ The radius is 1
∴ r = 1
∴ The equation is ⇒ (x - 0)² + (y - 2)² = 1²
∴ The equation is ⇒ x² + (y - 2)² = 1
∵ A circle represents the graph of a relation
∴ The equation determine a relation between x and y
* Lets prove that x=g(y)
- To do that find x in terms of y by separate x in side and all other
in the other side
∵ x² + (y - 2)² = 1
- Subtract (y - 2)² from both sides
∴ x² = 1 - (y - 2)²
- Take square root for both sides
∴ x = ±
∴ x = g(y)
* Lets prove that y=h(x)
- To do that find y in terms of x by separate y in side and all other
in the other side
∵ x² + (y - 2)² = 1
- Subtract x² from both sides
∴ (y - 2)² = 1 - x²
- Take square root for both sides
∴ y - 2 = ±
- Add 2 for both sides
∴ y = ±
∴ y = h(x)
- In the function x = ±
∵ ≥ 0
∴ 1 - (y - 2)² ≥ 0
- Add (y - 2)² to both sides
∴ 1 ≥ (y - 2)²
- Take √ for both sides
∴ 1 ≥ y - 2 ≥ -1
- Add 2 for both sides
∴ 3 ≥ y ≥ 1
∴ The domain is 1 ≤ y ≤ 3
- In the function y = ±
∵ ≥ 0
∴ 1 - x² ≥ 0
- Add x² for both sides
∴ 1 ≥ x²
- Take √ for both sides
∴ 1 ≥ x ≥ -1
∴ The domain is -1 ≤ x ≤ 1
* The graphs of these two function are half circle with center (0 , 2)
* All of the points on the circle that have distance 1 from point (0 , 2)
Answer:
c) 6x + y = -52 is required equation perpendicular to the given equation.
Step-by-step explanation:
If the equation is of the form : y = mx + C.
Here m = slope of the equation.
Two equations are said to be perpendicular if the product of their respective slopes is -1.
Here, equation 1 : -x + 6y = -12
or, 6y = -12 + x
or, y = (x/6) - 2
⇒Slope of line 1 = (1/6)
Now, for equation 2 to be perpendicular:
Check for each equation:
a. x + 6y = -67 ⇒ 6y = -67 - x
or, y = (-x/6) - (67/6) ⇒Slope of line 2 = (-1/6)
but
b. x - 6y = -52 ⇒ -6y = -52 - x
or, y = (x/6) + (52/6) ⇒Slope of line 2 = (1/6)
but
c. 6x + y = -52
or, y =y = -52 - 6x ⇒Slope of line 2 = (-6)
Hence, 6x + y = -52 is required equation 2.
d. 6x - y = 52 ⇒ -y = 52 - 6x
or, y = 6x - 52 ⇒Slope of line 2 = (6)
but
Hence, 6x + y = -52 is the only required equation .