Find the intervals on which the function is increasing, decreasing, concave up, or concave down and find any local extreme values and inflection points.
So I know how to do the above, but not with piece wise functions. Someone help me out?
Y={3-x^2, x<0 and x^2+1, x>=0}
I found the 1st and 2nd derivative for both, but I cannot set them to zero or anything because the 2nd derivative is a constant; y''=2 and y''=-2. I could graph this but it says to use algebra/analytic methods.
Thanks
clearly y is discontinuous at x=0, since
lim(x->0-) = 3
lim(x->0+) = 1
y'=-2x for x<0
y'=2x for x>0
y" = -2 for x<0
y" = +2 for x>0
so, y is increasing everywhere, since y'>0 everywhere it is defined.
y is concave down for x<0
y is concave up for y>0
Geez - just sketch the graphs, and all will become clear.
A useful tool is desmos.com, which can do piecewise graphs:
https://www.desmos.com/calculator/cqhgzeu7sg
Got it. Thank you for clearing that up!
To analyze the intervals of the piecewise function and determine the increasing, decreasing, concave up, and concave down intervals, as well as the local extreme values and inflection points, you can follow these steps:
Step 1: Find the critical points
Differentiate both pieces of the function separately to find the derivatives:
For x < 0:
f(x) = 3 - x^2
f'(x) = -2x
Set -2x = 0 to find the critical point:
-2x = 0
x = 0
For x ≥ 0:
f(x) = x^2 + 1
f'(x) = 2x
Set 2x = 0 to find the critical point:
2x = 0
x = 0
So, the critical point is x = 0.
Step 2: Analyze the intervals of increasing and decreasing
To determine where the function is increasing or decreasing, evaluate the sign of the derivatives in each interval. Consider x < 0 and x ≥ 0 separately:
For x < 0:
Since f'(x) = -2x, when x < 0, f'(x) is positive. This means that the function is increasing on the interval x < 0.
For x ≥ 0:
Since f'(x) = 2x, when x ≥ 0, f'(x) is positive. Thus, the function is also increasing on the interval x ≥ 0.
Therefore, the function is increasing on the entire domain (-∞, ∞).
Step 3: Analyze the intervals of concavity
To determine where the function is concave up or concave down, evaluate the sign of the second derivative in each interval:
For x < 0:
The second derivative is a constant, y'' = 2. Since 2 > 0, the function is concave up on the interval x < 0.
For x ≥ 0:
The second derivative is a constant, y'' = -2. Since -2 < 0, the function is concave down on the interval x ≥ 0.
Step 4: Find the local extreme values
To find the local extreme values, we need to evaluate the function at the critical point x = 0 and compare it to the function values nearby.
For x < 0:
Evaluate the function at x = 0 to find the corresponding y-value:
f(0) = 3 - (0)^2 = 3
Check nearby values:
f(-1) = 3 - (-1)^2 = 2
f(-2) = 3 - (-2)^2 = -1
From these calculations, we can see that the function has a local maximum at x = 0 since f(0) = 3 is greater than both f(-1) = 2 and f(-2) = -1.
For x ≥ 0:
Evaluate the function at x = 0:
f(0) = (0)^2 + 1 = 1
Check nearby values:
f(1) = (1)^2 + 1 = 2
f(2) = (2)^2 + 1 = 5
From these calculations, we can see that the function has a local minimum at x = 0 since f(0) = 1 is less than both f(1) = 2 and f(2) = 5.
Step 5: Find the inflection points
Since the concavity changes at x = 0, it is an inflection point. Evaluate the function beyond x = 0 and x < 0 to determine the behavior around the inflection point:
For x < 0:
Evaluate the function at x = -1:
f(-1) = 3 - (-1)^2 = 2
f(-2) = 3 - (-2)^2 = -1
For x ≥ 0:
Evaluate the function at x = 1:
f(1) = (1)^2 + 1 = 2
f(2) = (2)^2 + 1 = 5
From these calculations, we can see that the function changes concavity around x = 0.
In summary:
- The function is increasing on the entire domain (-∞, ∞).
- The function is concave up for x < 0 and concave down for x ≥ 0.
- The function has a local maximum at x = 0.
- The function has a local minimum at x = 0.
- The function has an inflection point at x = 0.
To find the intervals on which the piecewise function is increasing, decreasing, concave up, or concave down, as well as any local extreme values and inflection points, we can approach it by considering the two separate cases for the function:
Case 1: x < 0
In this case, the function is given by y = 3 - x^2. Let's find the first and second derivatives.
First derivative:
y' = d/dx (3 - x^2)
= 0 - d/dx(x^2)
= -2x
Second derivative:
y'' = d/dx (-2x)
= -2
Since the second derivative is negative, y'' = -2 in this interval, we conclude that the function is concave down in the interval x < 0.
To determine the intervals of increasing and decreasing, we look at the sign of the first derivative:
When x < 0, y' = -2x. Since the coefficient of x is negative, we know that the function is decreasing in the interval x < 0.
Case 2: x ≥ 0
In this case, the function is given by y = x^2 + 1. Let's find the first and second derivatives.
First derivative:
y' = d/dx (x^2 + 1)
= 2x + 0
= 2x
Second derivative:
y'' = d/dx (2x)
= 2
Since the second derivative is positive, y'' = 2 in this interval, we conclude that the function is concave up in the interval x ≥ 0.
To determine the intervals of increasing and decreasing, we look at the sign of the first derivative:
When x ≥ 0, y' = 2x. Since the coefficient of x is positive, we know that the function is increasing in the interval x ≥ 0.
In summary:
1. For x < 0, the function is decreasing and concave down.
2. For x ≥ 0, the function is increasing and concave up.
Now, let's find any local extreme values and inflection points:
Local extreme values:
Since the function is decreasing in the interval x < 0, there can be a local maximum but not a local minimum.
Inflection points:
Since the function is concave down in the interval x < 0 and concave up in the interval x ≥ 0, there could potentially be an inflection point at x = 0 since it is the point where the concavity changes. To confirm, let's check the signs of the second derivative on both sides of x = 0:
For x < 0, y'' = -2, which is negative (concave down).
For x > 0, y'' = 2, which is positive (concave up).
At x = 0, the concavity changes from concave down to concave up. Therefore, x = 0 is an inflection point.
In summary:
- The function has a local maximum value at x = 0.
- The function has an inflection point at x = 0.