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|

assuming you just need to confirm your answers, I'll say

none
|x+.5|>1.5
monotone over any interval not including the x-values -2,-.5,1
up: x<-2 and x>1
down: -2<x<1
no inflection

http://www.wolframalpha.com/input/?i=|x^2+%2Bx-2|

given the student's use of strange exponent notation it might be

|x^2 + 1/x^2 |

http://www.wolframalpha.com/input/?i=|x^2+%2Bx^-2|

To find the asymptote, interval of monotonicity, critical points, local extreme points, intervals of concavity, and inflection points of the function f(x) = |x^2 + x - 2|, we will follow these steps:

Step 1: Determine the domain of the function.

Since the function contains an absolute value, the expression inside the absolute value must be non-negative. Solve the inequality x^2 + x - 2 ≥ 0 to find the domain of the function.

x^2 + x - 2 ≥ 0
(x - 1)(x + 2) ≥ 0

We find that the function is non-negative when x ≤ -2 or x ≥ 1.

Thus, the domain of the function f(x) = |x^2 + x - 2| is (-∞, -2] ∪ [1, ∞).

Step 2: Find the asymptotes.

Since there are no rational terms in the function, there are no vertical asymptotes. However, there may be horizontal or slant asymptotes.

To determine if there are any asymptotes, we need to analyze the behavior of the function as x approaches positive and negative infinity.

As x approaches infinity, f(x) = |x^2 + x - 2| will take the form f(x) = |x^2| = x^2.

Therefore, as x approaches positive infinity, f(x) approaches infinity.

As x approaches negative infinity, f(x) = |x^2 + x - 2| will take the form f(x) = |x^2| = x^2.

Therefore, as x approaches negative infinity, f(x) approaches infinity.

Since f(x) approaches infinity for both positive and negative infinity, there are no horizontal or slant asymptotes.

Step 3: Find the critical points.

To find the critical points, we need to find the values of x where the derivative of the function is either equal to zero or does not exist.

The function f(x) = |x^2 + x - 2| is not differentiable at x = -2 and x = 1 (since the absolute value function has a sharp corner at these points).

Therefore, the critical points are x = -2 and x = 1.

Step 4: Find the local extreme points.

To find the local extreme points, we need to find the values of x for which f'(x) = 0 or does not exist.

Since f'(x) does not exist at x = -2 and x = 1, we need to examine the behavior of f(x) around these points.

When x < -2, f(x) = -(x^2 + x - 2).

When -2 < x < 1, f(x) = x^2 + x - 2.

When x > 1, f(x) = x^2 + x - 2.

From this analysis, we can see that f(x) has a local minimum at x = -2 and a local maximum at x = 1.

Step 5: Find the intervals of concavity and inflection points.

To find the intervals of concavity and inflection points, we need to find where the second derivative of the function is equal to zero or does not exist.

The second derivative is given by f''(x) = 2 for all x.

Since the second derivative is a constant and non-zero, there are no inflection points and the concavity does not change.

Step 6: Sketch the graph.

From the analysis above, we have the following information:

- Domain: (-∞, -2] ∪ [1, ∞)
- Asymptotes: None
- Critical points: x = -2, x = 1
- Local extreme points: Local minimum at x = -2, local maximum at x = 1
- Intervals of concavity: Concave up for all x
- Inflection points: None

We can now sketch the graph of f(x) = |x^2 + x - 2| using this information. However, without specific values for x, we cannot determine exact y-values.

To find the asymptote, interval of monotonicity, critical points, local extreme points, intervals of concavity, and inflection points of the function f(x) = |x^2 +x-2|, let's go step by step.

1. Asymptote: Since the function is in absolute value form, it doesn't have any vertical asymptotes. However, it could have horizontal asymptotes. To find them, we need to take the limit of f(x) as x goes towards positive and negative infinity.

As x approaches positive infinity, f(x) approaches |x^2 + x - 2| which is equal to x^2 + x - 2.

As x^2 + x - 2 has a leading term of x^2, the limit of f(x) as x approaches positive infinity is positive infinity. So, there is no horizontal asymptote.

Similarly, as x approaches negative infinity, f(x) approaches |x^2 + x - 2| which is equal to x^2 + x - 2.

Again, as x^2 + x - 2 has a leading term of x^2, the limit of f(x) as x approaches negative infinity is positive infinity. So, there is no horizontal asymptote.

Therefore, there are no asymptotes for this function.

2. Interval of monotonicity: To determine the intervals where the function is increasing or decreasing, we need to find the critical points and test the function in each interval.

Let's find the critical points by setting the derivative equal to zero and solving for x:

f'(x) = 0
|2x + 1| - (2x + 1) = 0

When 2x + 1 = 0, we get x = -1/2.

This is the only critical point of the function.

Now, let's test the function in different intervals:
For x < -1/2, we can pick x = -2 as a test point:
f(-2) = |-2^2 - 2 - 2| = |4 - 2 - 2| = 0.

For -1/2 < x < 0, we can pick x = -1 as a test point:
f(-1) = |-1^2 - 1 - 2| = |1 - 1 - 2| = 0.

For 0 < x, we can pick x = 1 as a test point:
f(1) = |1^2 + 1 - 2| = |1 + 1 - 2| = 0.

From this analysis, we can see that f(x) is equal to zero for x < -1/2, -1/2 < x < 0, and x > 0. In these intervals, the function doesn't increase or decrease since it is always zero. However, since f(x) changes from negative to positive at x = -1/2, this indicates a relative minimum at x = -1/2.

3. Local extreme points: As we found before, the only relative minimum is at x = -1/2.

4. Intervals of concavity: To find the intervals of concavity, we need to determine the second derivative and find where it is positive or negative.

f''(x) = 2 - 2/x^2

To find the intervals of concavity, we need to solve f''(x) = 0:

2 - 2/x^2 = 0
2 = 2/x^2
x^2 = 1
x = ±1

This gives us the critical points x = 1 and x = -1.

Testing intervals, we find that f''(x) > 0 for x < -1 and 0 < x < 1, indicating the function is concave up in these intervals. f''(x) < 0 for -1 < x < 0 and x > 1, indicating the function is concave down in these intervals.

5. Inflection point: From our analysis, we find that there are inflection points at x = -1 and x = 1 since the concavity changes at these points.

Finally, let's sketch the graph of the function f(x) = |x^2 + x - 2|:

The graph has a relative minimum at x = -1/2, and inflection points at x = -1 and x = 1. Since the function is always zero in the intervals x < -1/2, -1/2 < x < 0, and x > 0, the graph will touch the x-axis at these points. The function will be positive for x < -1/2 and x > 0, and negative for -1/2 < x < 0.

Overall, the graph will have a "V" shape with the vertex at (x, y) = (-1/2, 0), and it will touch the x-axis at x = -1 and x = 1.

Note that the shape of the graph might look different depending on the scale, so it's always helpful to plot a few more points and consider the behavior of the function near critical points.

I hope this explanation helps! Let me know if you have any further questions.