Option (A) is correct.
Step-by-step explanation: The given function is
We are to select the interval in which the above function is decreasing.
For the interval
So, f(x) is decreasing in this interval.
For the interval ,
So, f(x) isnot decreasing in this interval.
For the interval
So, f(x) is not decreasing in this interval.
For the interval
So, f(x) is not decreasing.
Thus, (A) is the correct option.
A. The extreme value theorem can apply to a piece wise function
C. An extreme value of a continuous function can be at one of the endpoints of the given closed interval
E. Given the function f(x)=x^2, the extreme value theorem guarantees the existence of a minimum and maximum value of the interval
I just took it on Edge and these were the correct answers !
answer is b iam sure this is correct
im not quite sure if this is right but i got 5.6 oisutuve
Increasing on it's domain because the slope is positive.
The domain and range are both all real numbers, also known as
All domain really means is what numbers can you plug in and you get number back from your function.
I should be able to plug in any number into 3x+2 and result in a number. There are no restrictions for x on 3x+2.
The domain is all real numbers.
In interval notation that is .
Now the range is the set of numbers that get hit by y=3x+2.
Well y=3x+2 is a linear function that is increasing. I know it is increasing because the slope is positive 3. I wrote out the positive part because that is the item you focus on in a linear equation to determine if is increasing or decreasing.
If slope is positive, then the line is increasing.
If slope is negative, then the line is decreasing.
So y=3x+2 hits all values of y because it is increasing forever. The range is all real numbers. In interval notation that is .
It's the open interval. The function increases over any interval on which its derivative is positive. We have
and for . So is increasing on , or .
To find concavity you must take the second derivative.
As you would to find your local maximums and minimums (critical points) in the first derivative by setting y' = 0, to find points of inflection you set acceleration, y" = 0.
Now that you know where the point in which the function is neither concave up or concave down (at the points of inflection) plug x-values between them into the second derivative for x. If y" is positive between those particular points will be concave up and if y" is negative it will be concave down between that interval.
For a better understanding you might find a good video on Youtube explaining this if you search "Points of Inflections" or "Concavity of a function".
(just did this question & got it right)
From the intervals shown in the question, the function decreases in the interval from x = π / 2 to x = π.
That is, π / 2 <x <π.
If, for example, we replace a value within this interval in the function and then replace a new value greater than the first, we can see that the value of f decreases.
If we substitute x = 2, then f = 2,615.
Then we substitute x = 2.2 and we have for this case that f = 1,229.
Finally 2,615 <1,229. This shows that in this interval the function is decreasing as x increases.
Below is a graph of the function f (x) = 4cos (2x-π) for the mentioned interval.
1. according to the condition
2. for more details see the attached graph.