F(x) is a polynomial with real coefficients. Hence for any real numbers x, ...
• f(x) is a real-valued function
• f(x) is continuous on any closed interval of real numbers, such as [4, 6]
• f(x) is differentiable on any open interval of real numbers, such as (4, 6)
The conclusion is that there exists some "c" such that f'(c) = (f(6) -f(4)/(6 - 4).
The slope of interest is
m = (f(6) -f(4))/(6 -4) = (-44 -(-44))/2 = 0
The slope f'(x) is -20 +4x. It will be zero where
0 = -20 +4x
20 = 4x
5 = x
So, f'(5) = 0 = m
The point "c" of interest is
c = {5}
Answer : (0,3) U (3, ∞)
We need to find the interval(s) over which the function is decreasing
We look at the intervals of x where the graph is decreasing
Decreasing interval of the graph is the part where the graph goes down
We look at the part where the graph changes direction
In the middle inverted U shaped graph , the graph changes direction and start decreasing at x=0
the graph is decreasing till x=3
so first decreasing interval is x=0 to 3. we write it as (0,3)
After x=3 the graph starts decreasing and it goes to infinity
So the second decreasing interval is 3 to infinity . We write it as (3, ∞)
Now combine both intervals by a 'U' union
(0,3) U (3, ∞)
Answer:
7
Step-by-step explanation:
Simplify the following:
13 (-5) - 8 (-9)
13 (-5) = -65:
-65 - 8 (-9)
-8 (-9) = 72:
-65 + 72
-65 + 72 = 7:
Answer: 7