Find the critical point and the interval on which the given function is increasing or decreasing, and apply the First Derivative Test to the critical point. Let
f(x)=4/x^2+1
Critical Point =
Is f a maximum or minumum at the critical point?
The interval on the left of the critical point is:
On this interval, f is_______ while f′ is ______
The interval on the right of the critical point is:
On this interval, f is_______while f′ is__________
To find the critical point of the function f(x), we need to find the value(s) of x where the derivative of f(x) is equal to zero or undefined.
Let's start by finding the derivative of f(x):
f(x) = 4/(x^2 + 1)
Using the Quotient Rule, we can differentiate the function:
f'(x) = [4'(x)(x^2 + 1) - 4(x^2 + 1)'(x)] / (x^2 + 1)^2
= [0(x^2 + 1) - 4(2x)] / (x^2 + 1)^2
= -8x / (x^2 + 1)^2
Now, we need to find the critical point(s) by setting the derivative equal to zero and solving for x:
-8x / (x^2 + 1)^2 = 0
Since the numerator is equal to zero, we have:
-8x = 0
x = 0
Therefore, the critical point of the function f(x) is x = 0.
To determine if f(x) has a maximum or minimum at the critical point, we can use the First Derivative Test.
The First Derivative Test states that if the derivative changes sign at a critical point, then the function has a relative maximum or minimum at that point.
To apply the test, we can choose a point to the left and right of the critical point, and determine the signs of both f(x) and f'(x) at those points.
Let's choose x = -1 as the point to the left of the critical point, and x = 1 as the point to the right.
For x = -1:
On this interval, f(x) is positive while f'(x) is negative.
For x = 1:
On this interval, f(x) is positive while f'(x) is positive.
So, based on the First Derivative Test, at the critical point x = 0, f(x) changes from decreasing (as f'(x) is negative) to increasing (as f'(x) becomes positive). Therefore, we have a local minimum at x = 0.
The interval on the left of the critical point is (-∞, 0). On this interval, f is positive while f' is negative.
The interval on the right of the critical point is (0, ∞). On this interval, f is positive while f' is positive.
To find the critical point of the given function f(x) = 4/x^2+1, we need to find where the derivative of the function equals zero or is undefined.
First, let's find the derivative of the function f(x) = 4/x^2+1. We can use the quotient rule to find the derivative:
f'(x) = [(4)'(x^2+1) - (4)(x^2+1)'] / (x^2+1)^2
f'(x) = (0 - 8x) / (x^2+1)^2
f'(x) = -8x / (x^2+1)^2
To find the critical point, we set f'(x) equal to zero and solve for x:
-8x / (x^2+1)^2 = 0
Since the numerator is always zero, the critical point occurs when the denominator equals zero:
x^2+1 = 0
Solving this equation, we find that there are no real solutions since x^2 can't be negative. Therefore, there are no critical points for this function.
Since there are no critical points, there is no maximum or minimum value to determine.
Now, let's discuss the intervals of the function to determine when it is increasing or decreasing.
First, consider the interval on the left of the critical point. Since there is no critical point, we can take any arbitrary number from the interval, such as x = -1. Plugging this value into the derivative:
f'(-1) = -8(-1) / ((-1)^2+1)^2
f'(-1) = 8 / 4
f'(-1) = 2
Since the derivative is positive (f'(-1) = 2 > 0), we can conclude that on the interval to the left of the critical point, the function f(x) = 4/x^2+1 is increasing.
Next, consider the interval on the right of the critical point. Again, since there is no critical point, we can take any arbitrary number from the interval, such as x = 1. Plugging this value into the derivative:
f'(1) = -8(1) / ((1)^2+1)^2
f'(1) = -8 / 4
f'(1) = -2
Since the derivative is negative (f'(1) = -2 < 0), we can conclude that on the interval to the right of the critical point, the function f(x) = 4/x^2+1 is decreasing.
In summary:
- There are no critical points for the function f(x) = 4/x^2+1.
- Therefore, there is neither a maximum nor a minimum value.
- On the interval to the left of the critical point, the function is increasing.
- On the interval to the right of the critical point, the function is decreasing.
What, no ideas on any of your homework dump? I'll do this one, and the steps should make it easy for you to try the others and show what you did.
As you surely recall,
f(x) is increasing where f'(x) > 0
f(x) is decreasing where f'(x) < 0
f(x) is a max or min where f'(x) = 0
So, for your function, assuming the usual carelessness with parentheses, is
f(x) = 4/(x^2+1)
f'(x) = -8x/(x^2+1)^2
Now, the denominator is always positive, so
f'(x) > 0 where x < 0
f'(x) < 0 where x > 0
f'(x) = 0 where x = 0
so,
f(x) is increasing on (-∞,0)
f(x) is decreasing on (0,∞)
f(0) is a maximum, since the slope changes from + on the left to - on the right
there are many good online graphing sites. plot your function there, and see that the above is correct