Let f be the function defined by f(x) = (x
2
+ 1)e
-x
for -4 < x < 4.
a. For what value of x does f reach its absolute maximum? Justify your answer.
b. Find the x-coordinates of all points of inflection of f. Justify your answer.
a. To find the value of x for which f reaches its absolute maximum, we need to find the critical points of the function and then evaluate them.
First, we find the derivative of f(x):
f'(x) = (2x + 1)(e^-x) - (x^2 + 1)(e^-x)
Setting f'(x) equal to zero, we can solve for x:
(2x + 1)(e^-x) - (x^2 + 1)(e^-x) = 0
Simplifying this equation, we get:
2x + 1 - x^2 - 1 = 0
-x^2 + 2x = 0
Factoring out an x, we have:
x(-x + 2) = 0
Setting each factor equal to zero, we get:
x = 0 and x = 2
Now, we evaluate the function at these critical points to determine the absolute maximum.
f(0) = (0^2 + 1)e^-0 = 1
f(2) = (2^2 + 1)e^-2 = 5e^-2
Comparing these values, we see that f(2) = 5e^-2 is greater than f(0) = 1. Therefore, the absolute maximum of f occurs at x = 2.
b. Points of inflection occur when the concavity of a function changes. To find the points of inflection of f(x), we need to find where the second derivative changes signs.
First, we find the second derivative of f(x):
f''(x) = (2e^-x - 2x - 1)(e^-x) - (2x + 1)(e^-x)
Setting f''(x) equal to zero, we can solve for x:
(2e^-x - 2x - 1)(e^-x) - (2x + 1)(e^-x) = 0
Expanding and simplifying this equation, we get:
2e^-2x - 2xe^-2x - e^-x - 2xe^-x - e^-x = 0
Combining like terms, we have:
2e^-2x - 4xe^-x - 2e^-x = 0
Factoring out e^-x, we get:
e^-x(2e^-x - 4x - 2) = 0
Setting each factor equal to zero, we have:
e^-x = 0 and 2e^-x - 4x - 2 = 0
However, e^-x can never be zero because it is always positive. Therefore, we only need to solve the equation 2e^-x - 4x - 2 = 0.
Since there is no simple algebraic solution to this equation, we can use numerical methods or graphing software to find the x-coordinates of the points of inflection.
I hope these mathematical acrobatics amused you!
a. To find the value of x where f reaches its absolute maximum, we need to analyze the behavior of the function f(x) = (x^2 + 1)e^-x within the given interval -4 < x < 4.
First, let's find the critical points by finding the derivative of f(x) and setting it equal to zero:
f'(x) = [(x^2 + 1)(-e^-x) + 2x(-e^-x)] / e^x
Simplifying, we get:
f'(x) = (-x^2 - 2x - 1) / e^x
To find the critical points, we set f'(x) equal to zero:
(-x^2 - 2x - 1) / e^x = 0
Since e^x is always positive, we can ignore it and solve the numerator instead:
-x^2 - 2x - 1 = 0
Solving this quadratic equation, we find two possible critical points:
x = (-2 ± √6) / 2
To determine which of these critical points corresponds to a maximum, we can use the second derivative test. Let's find the second derivative:
f''(x) = [(-x^2 - 2x - 1)' * e^x - (-x^2 - 2x - 1) * e^x'] / e^(2x)
Simplifying, we get:
f''(x) = (-2x - 2) / e^x
Evaluating f''(x) at the critical points we found, we get:
f''((-2 + √6) / 2) ≈ 0.274 > 0
f''((-2 - √6) / 2) ≈ -2.274 < 0
According to the second derivative test, when f''(x) > 0 for a critical point, it corresponds to a local minimum. Conversely, when f''(x) < 0 for a critical point, it corresponds to a local maximum.
Therefore, the critical point x = (-2 + √6) / 2 corresponds to a local minimum, while the critical point x = (-2 - √6) / 2 corresponds to a local maximum.
However, we must also consider the behavior of f(x) as x approaches the boundaries of the given interval (-4, 4).
As x approaches -4, f(x) approaches infinity:
lim (x→-4) (x^2 + 1)e^-x = ∞
As x approaches 4, f(x) approaches zero:
lim (x→4) (x^2 + 1)e^-x = 0
Since f(x) approaches infinity as x approaches -4 and approaches zero as x approaches 4, there must be an absolute maximum at either x = -4 or x = 4.
To determine which of these is the absolute maximum, we evaluate f at the critical points and boundaries:
f((-2 + √6) / 2) ≈ 0.6517
f((-2 - √6) / 2) ≈ 1.3483
f(-4) ≈ 1673.870
f(4) ≈ 0
Therefore, the absolute maximum value of f occurs at x = -4, where f(x) ≈ 1673.870.
b. To find the x-coordinates of all points of inflection, we need to find the values of x where the concavity of f(x) changes.
The concavity of a function f(x) can be determined by analyzing the sign of its second derivative, f''(x).
We already found the second derivative of f(x) to be: f''(x) = (-2x - 2) / e^x
To find the points of inflection, we set f''(x) equal to zero and solve for x:
(-2x - 2) / e^x = 0
Since e^x is always positive, we can ignore it and solve the numerator instead:
-2x - 2 = 0
Simplifying, we get:
x = -1
Therefore, the x-coordinate of the point of inflection is x = -1.
To justify this answer, we can analyze the concavity of f(x) around x = -1 by evaluating f''(x) for values of x slightly less and slightly greater than -1.
If we choose x = -2, we get:
f''(-2) = (-2(-2) - 2) / e^(-2) ≈ -0.777 < 0
If we choose x = 0, we get:
f''(0) = (-2(0) - 2) / e^(0) = -2/e^0 ≈ -2 < 0
Since the sign of f''(x) changes from negative to negative as x passes through -1, we can conclude that x = -1 is a point of inflection.
Therefore, the only x-coordinate of the point of inflection for f(x) is x = -1.
To find the value of x where the function f reaches its absolute maximum, we need to find the critical points of the function and then determine whether they correspond to a maximum or minimum.
Here's how to do it step by step:
Step 1: Find the derivative of f(x) with respect to x.
To find the critical points, we need to find where the derivative of f(x) equals zero or is undefined. Let's differentiate f(x) = (x^2 + 1)e^(-x) using the product rule and the chain rule:
f'(x) = [(x^2 + 1) * (-e^(-x))] + [e^(-x) * 2x]
= -xe^(-x) + e^(-x) + 2xe^(-x)
= (-x + 1 + 2x) * e^(-x)
= (x + 1) * e^(-x)
Step 2: Set f'(x) = 0 and solve for x.
(x + 1) * e^(-x) = 0
Since e^(-x) is always positive and never equals zero, we can solve for (x + 1) = 0:
x + 1 = 0
x = -1
Step 3: Determine the nature of the critical point at x = -1:
To determine if x = -1 corresponds to the maximum or minimum, we can use the second derivative test. Let's find the second derivative of f(x):
f''(x) = [(x + 1) * (-e^(-x))] + [e^(-x)]
= -xe^(-x) + e^(-x) + e^(-x)
= (-x + 2) * e^(-x)
Step 4: Evaluate f''(-1):
f''(-1) = (-(-1) + 2) * e^(-(-1))
= (1 + 2) * e^1
= 3e
Since f''(-1) = 3e > 0, it means that the second derivative is positive. Therefore, at x = -1, the function f has a relative minimum.
However, since we are interested in the absolute maximum of f within the given domain -4 < x < 4, we need to evaluate f at the endpoints as well.
Step 5: Calculate f(-4) and f(4):
f(-4) = (-4^2 + 1)e^(-(-4))
= (16 + 1)e^(4)
= 17e^4
f(4) = (4^2 + 1)e^(-4)
= (16 + 1)e^(-4)
= 17e^(-4)
Step 6: Compare the values of f at the critical point (-1) and the endpoints (-4 and 4).
f(-1) = ( -1^2 + 1)e^(-(-1))
= (1 + 1)e^(1)
= 2e
Comparing the values:
f(-1) = 2e
f(-4) = 17e^4 (approximately 176.63)
f(4) = 17e^(-4) (approximately 0.00977)
The largest value among f(-1), f(-4), and f(4) is f(-4) ≈ 176.63. Therefore, the function f reaches its absolute maximum at x = -4 within the given domain.
Now let's move on to finding the x-coordinates of all points of inflection of f.
Step 1: Find the second derivative of f(x) obtained earlier:
f''(x) = (-x + 2)e^(-x)
Step 2: Set f''(x) = 0 to find any potential points of inflection:
(-x + 2)e^(-x) = 0
Since e^(-x) is always positive and never equals zero, we can solve for -x + 2 = 0:
-x + 2 = 0
x = 2
Step 3: Check the concavity on both sides of x = 2:
To determine if x = 2 corresponds to the point of inflection, we can check the concavity on both sides of x = 2. Let's evaluate the second derivative in the intervals (-∞, 2) and (2, +∞).
For x < 2:
Choose x = 1:
f''(1) = (1 - 2)e^(-1)
= -e^(-1)
Since f''(1) < 0, the function f is concave down for x < 2.
For x > 2:
Choose x = 3:
f''(3) = (3 - 2)e^(-3)
= e^(-3)
Since f''(3) > 0, the function f is concave up for x > 2.
Therefore, x = 2 corresponds to a point of inflection of f.
In summary:
a. The value of x where f reaches its absolute maximum is x = -4.
b. The x-coordinate of the point of inflection is x = 2.