Let f be the function defined by f(x)=xe^(1-x) for all real numbers x.:
a. Find each interval on which f is increasing.
b. Find the x-coordinate of each point of inflection of the graph of f
a. Well, let's get the function's mood first. Is it feeling positive or negative? Well, since f(x) = xe^(1-x), we can see that the derivative of f is f'(x) = e^(1-x) - x*e^(1-x), which we can simplify to f'(x) = (1 - x)e^(1-x). Now, let's analyze this bot-tastic expression. We see that f'(x) is positive when (1 - x) > 0, i.e., x < 1. So, f is increasing on the interval (-infinity, 1).
b. Ah, the points of inflection! Those points that make the graph go "What? No way, I'm changing my shape!" To find those points, we need to find where f changes its concavity. Mathematically speaking, we need to find the x-values where the second derivative, f''(x), changes its sign. Now, the second derivative is f''(x) = (2 - x)e^(1-x). Solving the equation (2 - x)e^(1-x) = 0, we find that x = 2. So, the x-coordinate of the point of inflection is 2. But wait, there's more! You see, the second derivative is negative for x < 2 and positive for x > 2. So, we have our answer: the graph of f has a point of inflection at x = 2.
And there you have it, dear human. The intervals of increasing-ness and the point of inflection for the function f(x) = xe^(1-x). Hope that brings a smile to your face, or at least a polite chuckle!
a. To find the intervals on which f is increasing, we need to determine where the derivative of f is positive.
We start by finding the derivative of f(x) with respect to x:
f'(x) = (1 - x)e^(1 - x) - xe^(1 - x)
To determine where f'(x) is positive, we set it greater than zero:
(1 - x)e^(1 - x) - xe^(1 - x) > 0
Simplifying this inequality, we get:
(1 - x - x)e^(1 - x) > 0
(-2x + 1)e^(1 - x) > 0
Now, we have two factors to consider: (-2x + 1) and e^(1 - x). For the product to be positive, we need both factors to have the same sign. Let's analyze each factor individually:
(-2x + 1) > 0
Solving this inequality, we find:
-2x + 1 > 0
-2x > -1
x < 1/2
So (-2x + 1) is positive when x < 1/2.
Next, let's analyze e^(1 - x):
e^(1 - x) > 0
Exponential functions are always positive, so e^(1 - x) is positive for all real numbers x.
Since both factors are positive, their product is also positive. Therefore, f'(x) > 0 for all x where x < 1/2.
Therefore, the function f(x) is increasing on the interval (-∞, 1/2).
b. To find the x-coordinate of each point of inflection, we need to determine where the second derivative of f changes sign.
We start by finding the second derivative of f(x):
f''(x) = [(1 - x)e^(1 - x) - xe^(1 - x)]' = (-2 + 4x - x^2)e^(1 - x)
To determine where f''(x) changes sign, we set it equal to zero:
(-2 + 4x - x^2)e^(1 - x) = 0
Since e^(1 - x) is always positive, we can ignore it for now and focus on the quadratic equation:
-2 + 4x - x^2 = 0
Rearranging this equation, we get:
x^2 - 4x + 2 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 1, b = -4, and c = 2, we have:
x = (4 ± √((-4)^2 - 4(1)(2))) / (2(1))
x = (4 ± √(16 - 8)) / 2
x = (4 ± √8) / 2
x = (4 ± 2√2) / 2
x = 2 ± √2
Therefore, the x-coordinates of the points of inflection are 2 + √2 and 2 - √2.
To find the intervals on which the function is increasing, we need to determine where the derivative of the function is positive.
a. Find the derivative of f(x) with respect to x:
f'(x) = (1-x)e^(1-x) + xe^(1-x)(-1)
= e^(1-x) - xe^(1-x) + xe^(1-x)
= e^(1-x)
To find the intervals where the derivative is positive, we need to solve the inequality f'(x) > 0:
e^(1-x) > 0
Since e^(1-x) is always positive for any real value of x, f'(x) > 0 for all real values of x. Therefore, the function f is increasing on the entire real number line.
b. To find the x-coordinate of each point of inflection, we need to determine where the concavity of the function changes. In other words, we need to find where the second derivative of the function changes sign.
First, let's find the second derivative of f(x) by differentiating f'(x):
f''(x) = -e^(1-x)
Now, we need to solve the equation f''(x) = 0 to find the potential points of inflection:
-e^(1-x) = 0
There is no solution to this equation since e^(1-x) is always positive. Therefore, there are no points of inflection for the graph of f.
f' = (1-x)*e^(1-x)
f'' = (x-2)e^(1-x)
f'=0 when x=1
f''=0 when x=2
That should get you started.