f(x) = e^x/1+e^x
find the horizontal and vertical assymptotes?
find the interval whr f is increasing?
Find the inflection point.
Find the interval where the function is concave up.
Find the interval where the function is concave down.
To find the horizontal and vertical asymptotes of the function f(x) = e^x / (1 + e^x), we can analyze the behavior of the function as x approaches positive or negative infinity.
1. Horizontal Asymptote:
As x approaches positive or negative infinity, the denominator (1 + e^x) tends to infinity, and the numerator (e^x) also tends to infinity. Therefore, the limit of the function as x approaches infinity (or negative infinity) is:
lim(x→±∞) f(x) = lim(x→±∞) e^x / (1 + e^x) = ∞ / ∞
To evaluate this type of limit, we can make use of L'Hôpital's Rule by taking the derivatives of the numerator and denominator with respect to x:
lim(x→±∞) e^x / (1 + e^x) = lim(x→±∞) (e^x)' / ((1 + e^x)')
Differentiating the numerator and denominator:
lim(x→±∞) e^x / (1 + e^x) = lim(x→±∞) e^x / e^x = 1
Therefore, the horizontal asymptote is y = 1.
2. Vertical Asymptote:
To find the vertical asymptote, we need to look for values of x that make the denominator zero. In this case, the denominator (1 + e^x) cannot be equal to zero because e^x is always positive and adding 1 to it will always result in a positive value. Hence, there are no vertical asymptotes for this function.
3. Increasing Intervals:
To determine where the function is increasing, we need to find the intervals where the derivative is positive. Let's find the derivative of f(x) first:
f(x) = e^x / (1 + e^x)
Using the quotient rule, we can differentiate f(x):
f'(x) = [(1 + e^x)(e^x)' - e^x(e^x)'] / (1 + e^x)^2
= [e^x + e^2x - e^2x] / (1 + e^x)^2
= e^x / (1 + e^x)^2
To determine the sign of the derivative, we look for values of x that make the numerator positive. Since e^x is always positive, the derivative is also positive for all values of x. Therefore, the function f(x) is increasing for all x in its domain.
4. Inflection Point:
To find the inflection point, we need to locate the x-value where the concavity changes. In this case, we need to find the x-value that makes the second derivative, f''(x), equal to zero.
Let's find the second derivative of f(x):
f'(x) = e^x / (1 + e^x)^2
Using the quotient rule, we differentiate f'(x):
f''(x) = [(1 + e^x)^2(e^x)' - e^x(2(1 + e^x)(e^x)')] / (1 + e^x)^4
= [e^x(1 + e^x)^2 - 2e^x(1 + e^x)(e^x)] / (1 + e^x)^4
= e^x(1 + e^x - 2e^x) / (1 + e^x)^4
= e^x(1 - e^x) / (1 + e^x)^3
To find the inflection point, we set f''(x) = 0:
e^x(1 - e^x) / (1 + e^x)^3 = 0
Since e^x is always positive and cannot be zero, we have:
1 - e^x = 0
e^x = 1
x = ln(1) = 0
Therefore, the inflection point occurs at x = 0.
5. Concave Up and Concave Down Intervals:
To determine where the function is concave up or concave down, we look at the sign of the second derivative, f''(x).
The second derivative, f''(x), can be factored as:
f''(x) = e^x(1 - e^x) / (1 + e^x)^3
To analyze the sign of f''(x), we examine the signs of each individual factor.
For e^x, since it is always positive, it does not affect the overall sign.
For (1 - e^x), the sign depends on the value of e^x.
- If e^x < 1, then (1 - e^x) > 0.
- If e^x > 1, then (1 - e^x) < 0.
Since we know the critical point is at x = 0, we test the intervals around this point to determine concavity:
For x < 0, e^x < 1, so (1 - e^x) > 0. Therefore, f''(x) > 0, which means the function is concave up.
For x > 0, e^x > 1, so (1 - e^x) < 0. Therefore, f''(x) < 0, which means the function is concave down.
Hence, the function f(x) is concave up for x < 0 and concave down for x > 0.