Find the asymptote, interval of monotonicity, critical points, the local extreme points, intervals of concavity and inflection point of the following functions. Sketch the graph.
a)f(x)=|x2 +x-2|
the graph can be seen here:
http://www.wolframalpha.com/input/?i=plot+y%3D|x^2+%2Bx-2|+for+-4+%3C%3D+x+%3C%3D+3
That should make it easy to answer the other questions. If not, come on back and show where you get stuck.
I think we went through this before.
note I think the somewhat unusual exponent notation may mean
| x^2 + x^-2| = | x^2 + 1/x^2 |
http://www.wolframalpha.com/input/?i=plot+y%3D|+x^2+%2B+1%2Fx^2+|
To find the asymptote, interval of monotonicity, critical points, local extreme points, intervals of concavity, and inflection point of the function f(x) = |x^2 + x - 2|, we can follow these steps:
1. Determine the domain:
Since we don't have any restrictions, the domain of f(x) is all real numbers.
2. Find the vertical asymptote:
The function f(x) = |x^2 + x - 2| is defined for all real numbers, so it doesn't have any vertical asymptotes.
3. Determine the interval of monotonicity:
To find the intervals of monotonicity, we need to find the critical points first. Since the function involves an absolute value, we need to consider both the positive and negative cases.
Setting the inside expression, x^2 + x - 2, equal to zero and solving for x, we have:
x^2 + x - 2 = 0
Factoring the quadratic equation, we get:
(x - 1)(x + 2) = 0
So, we have two critical points: x = 1 and x = -2. Now, we need to test the intervals created by these critical points.
For x < -2:
Pick a value less than -2, like x = -3. Substitute it into the original equation:
f(-3) = |-3^2 + (-3) - 2| = |9 - 3 - 2| = |4| = 4
For x between -2 and 1:
Pick a value between -2 and 1, like x = 0. Substitute it into the original equation:
f(0) = |0^2 + 0 - 2| = |-2| = 2
For x > 1:
Pick a value greater than 1, like x = 2. Substitute it into the original equation:
f(2) = |2^2 + 2 - 2| = |4| = 4
From these calculations, we can see that the function increases for x < -2, decreases for -2 < x < 1, and increases again for x > 1. Therefore, the interval of monotonicity is (-∞, -2) U (1, ∞).
4. Find the critical points:
We already found the critical points in step 3: x = 1 and x = -2.
5. Determine the local extreme points:
To find the local extreme points, we need to analyze the behavior of the function around the critical points.
For x = -2:
Substitute x = -2 into the original equation:
f(-2) = |-2^2 + (-2) - 2| = |4 - 2 - 2| = 0
For x = 1:
Substitute x = 1 into the original equation:
f(1) = |1^2 + 1 - 2| = |1 + 1 - 2| = 0
Both points x = -2 and x = 1 correspond to local minimum points since the function reaches a minimum value of 0 at these points.
6. Find the intervals of concavity:
To find the intervals of concavity, we need to examine the second derivative. However, since the absolute value function is not differentiable at certain points, we cannot find the concavity intervals or the inflection point.
7. Sketch the graph:
Using the obtained information, we can sketch the graph of f(x) = |x^2 + x - 2|. The graph will have a local minimum at x = -2, increase, reach another local minimum at x = 1, and then continue increasing. It won't have any vertical asymptotes or inflection points.
Note: If you need a visual representation of the graph, you can graph the function using graphing software, a calculator, or an online graphing tool.