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 of each.
A) f(x)=x^2e^x
B) f(x)=|x^2 +x-2|
C) f(x)= x^2 -6x/(x+1)^2
A) Let's analyze the function f(x) = x^2 * e^x.
1. Asymptote: To find the asymptote, we need to check the limits as x approaches positive and negative infinity.
- As x approaches negative infinity, e^x approaches zero, so the term x^2 * e^x approaches zero. Thus, there is no asymptote as x approaches negative infinity.
- As x approaches positive infinity, both x^2 and e^x approach infinity. Therefore, there is no asymptote as x approaches positive infinity.
2. Interval of monotonicity: To find the interval of monotonicity, we need to find where the derivative is positive or negative.
- First, find the derivative of f(x): f'(x) = (2x * e^x) + (x^2 * e^x)
- Setting f'(x) > 0, we have (2x * e^x) + (x^2 * e^x) > 0
- Factoring out e^x, we get e^x(2x + x^2) > 0
- Now, we check each interval to determine the sign of each factor:
a) For x < -2, both e^x and (2x + x^2) are positive. Therefore, f'(x) > 0.
b) For -2 < x < 0, e^x is positive, but (2x + x^2) is negative. Therefore, f'(x) < 0.
c) For x > 0, both e^x and (2x + x^2) are positive. Therefore, f'(x) > 0.
- Thus, the function f(x) is increasing on (-∞, -2) U (0, +∞) and decreasing on (-2, 0).
3. Critical points: Critical points occur where the derivative is either zero or undefined.
- Setting f'(x) = 0, we have (2x * e^x) + (x^2 * e^x) = 0
- Factoring out the common factor of e^x, we get e^x(2x + x^2) = 0
- From this equation, we have two possibilities:
a) e^x = 0, which has no solutions since e^x is always positive.
b) (2x + x^2) = 0, which can be solved to find the critical points x = -2 and x = 0.
4. Local extreme points: To find local extreme points, we check the behavior on each side of the critical points.
- For x < -2, the function is increasing. So, there is no local minimum or maximum.
- At x = -2, the function changes from increasing to decreasing, indicating a local maximum.
- For -2 < x < 0, the function is decreasing. So, there is no local minimum or maximum.
- At x = 0, the function changes from decreasing to increasing, indicating a local minimum.
- For x > 0, the function is increasing. So, there is no local minimum or maximum.
5. Intervals of concavity: To find the intervals of concavity, we need to analyze the second derivative.
- Find the second derivative of f(x): f''(x) = (2 * e^x) + (4x * e^x) + (x^2 * e^x)
- To determine the signs of the second derivative, we need to analyze each factor:
a) For x < -2, all three factors are positive. Therefore, f''(x) > 0.
b) For -2 < x < 0, e^x is positive, while (2 * e^x) + (4x * e^x) + (x^2 * e^x) is negative. Thus, f''(x) < 0.
c) For x > 0, all three factors are positive. Therefore, f''(x) > 0.
- The function f(x) is concave up on (-∞, -2) U (0, +∞) and concave down on (-2, 0).
6. Inflection point: Inflection points occur where the concavity changes.
- At x = -2, the concavity changes from concave up to concave down, indicating an inflection point.
Now, you can sketch the graph of the function f(x) = x^2 * e^x using the information you have gathered.