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)=x2ex
B) f(x)=|x2 +x-2|
C) f(x)= x2 -6x
(x+1)2
A) f(x) = x^2 e^x , notice how I indicated exponents?
f ' (x) = x^2 e^x + 2x e^x
= e^x(x^2 + 2x)
f'' (x) = x^2 e^x + 2x e^x + 2x e^x + 2 e^x
= e^x (x^2+ 4x + 2)
here is a graph of your first function
http://www.wolframalpha.com/input/?i=y+%3D+x%5E2+e%5Ex
type in your other equations to get the graphs of them
for your first one:
there is no vertical asymtote, as x ---> large, f(x) --> large
but there is a horizontal asymptote of y = 0
as x ----> -large, f(x) becomes smaller and ---> 0
set f ' (x) = 0
e^x(x^2 + 2x)
e^x = 0 ---> no solution
or x^2 + 2x = 0
x(x+2) = 0
x = 0 or x = -2
close up of above result:
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+e%5Ex%2C++-3+%3C+x+%3C+1
f(0) = 0
f(-2) = 4e^-2 = appr .54
from a graph we have a local max at (-2, .54) and a local min at (0,0)
set f '' (x) = 0
e^x (x^2 + 4x + 2) = 0
e^x= 0 , no solution
or x2+ 4x + 2 = 0
x^2+ 4x + 4 = -2 + 4 , I am using completing the square
(x+2)^2 = 2
x+2 = ± √2
x = -2 ± √2
So two points of inflection, as seen in
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+e%5Ex%2C++-8+%3C+x+%3C+1
All you have to do is enter your other equations in this fantastic webpage.
To find the asymptote, interval of monotonicity, critical points, local extreme points, intervals of concavity, and inflection points for each function and sketch their graphs, we will go step by step. Let's start with function A.
A) f(x) = x^2 * e^x
1. Asymptote:
To find the asymptote, we need to check the behavior of the function as x approaches positive or negative infinity. In this case, as x approaches positive or negative infinity, the exponential term e^x will grow much faster than x^2, so there is no horizontal asymptote.
However, we can check if there is any vertical asymptote by examining the behavior of the denominator of the function. In this case, there is no denominator, so there are no vertical asymptotes either.
2. Interval of Monotonicity:
To find the interval of monotonicity, we need to determine where the function is increasing or decreasing. We can do this by finding the critical points and checking the sign of the derivative.
Let's take the derivative of f(x) with respect to x:
f'(x) = 2x * e^x + x^2 * e^x = (2x + x^2) * e^x
Setting f'(x) = 0 to find the critical points:
(2x + x^2) * e^x = 0
From this equation, we can see that e^x cannot be zero (as it is always positive), so we can solve for the polynomial part:
2x + x^2 = 0
This quadratic equation has two solutions:
x = 0 and x = -2
Thus, the critical points are x = 0 and x = -2.
To determine the intervals of monotonicity, we need to analyze the signs of f'(x) on different intervals separated by the critical points:
On the interval (-∞, -2), we pick a value to the left of -2, say x = -3:
f'(-3) = (2*(-3) + (-3)^2) * e^(-3) = -3 * e^(-3) < 0
So f'(x) is negative on (-∞, -2).
On the interval (-2, 0), we pick a value between -2 and 0, say x = -1:
f'(-1) = (2*(-1) + (-1)^2) * e^(-1) = (-1) * e^(-1) < 0
So f'(x) is negative on (-2, 0).
On the interval (0, ∞), we pick a value to the right of 0, say x = 1:
f'(1) = (2*1 + 1^2) * e^1 = 3 * e > 0
So f'(x) is positive on (0, ∞).
From this information, we can determine the interval of monotonicity as follows:
Focusing on the function f(x), we see that as x increases from negative infinity to -2, f(x) is decreasing. Between -2 and 0, f(x) is still decreasing. Finally, as x increases from 0 to positive infinity, f(x) is increasing.
3. Critical Points and Local Extreme Points:
The critical points are the values of x where the derivative is either zero or undefined. In this case, we already found the critical points to be x = 0 and x = -2. To determine whether these critical points are local maximum or minimum points, we can use the second derivative test:
Taking the derivative of f'(x):
f''(x) = (2 + 2x) * e^x
Evaluate f''(x) at x = 0:
f''(0) = (2 + 2(0)) * e^0 = 2 > 0
Since f''(0) is positive, the critical point x = 0 corresponds to a local minimum.
Evaluate f''(x) at x = -2:
f''(-2) = (2 + 2(-2)) * e^(-2) = -4 * e^(-2) < 0
Since f''(-2) is negative, the critical point x = -2 corresponds to a local maximum.
Therefore, the function has a local minimum at x = 0 and a local maximum at x = -2.
4. Intervals of Concavity and Inflection Points:
To find the intervals of concavity and the inflection points, we need to analyze the behavior of the second derivative, f''(x).
On the interval (-∞, -2), we pick a value to the left of -2, say x = -3:
f''(-3) = (2 + 2*(-3)) * e^(-3) = -4 * e^(-3) < 0
So f''(x) is negative on (-∞, -2).
On the interval (-2, ∞), we pick a value to the right of -2, say x = 1:
f''(1) = (2 + 2(1)) * e = 4 * e > 0
So f''(x) is positive on (-2, ∞).
From this information, we can determine the intervals of concavity as follows:
Between negative infinity and -2, the function is concave downward. Between -2 and infinity, the function is concave upward.
To find the inflection point(s), we need to see where the concavity changes.
Since we have determined the interval of concavity and the critical points, we can conclude that the inflection point is at x = -2.
5. Sketching the Graph:
Based on the information we found, we can sketch the graph of the function f(x) = x^2 * e^x. The graph will have no asymptotes and will have a local minimum at x = 0 and a local maximum at x = -2. The graph will be concave downward in the interval (-∞, -2) and concave upward in the interval (-2, ∞). The inflection point is at x = -2.
Please note that it is not feasible to draw a graph using text, so I would recommend using graphing software or plotting points to draw the graph accurately.