Consider the function below. (Round the answers to two decimal places. If you need to use - or , enter -INFINITY or INFINITY.)
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 have a vertical asymptote, 1 + e^x has to be zero, or
e^x = -1
No value of x makes that statement true, so there is no vertical asymptote.
for x --->∞ , f(x) approaches 1
for x ---> -∞, f(x) approaches 0
so for large +x's, the horizontal asymptote is y = 1
for large -x's the horizontal asymptote is y = 0
f'(x) = [(1+e^x)(e^x) - e^x(e^x)]/(1+e^x)^2
= e^x(1 + 2e^x)/(1+e^x)^2
This will always be positive for any value of x, so the function is increasing for all values of x
so would the inflection points be (0,1)?
To find the horizontal asymptote, we need to determine the behavior of the function as x approaches positive infinity and negative infinity.
As x approaches negative infinity, the denominator of the function approaches 1+e^x, which is positive because e^x is always positive. Therefore, as x approaches negative infinity, the function approaches 0.
As x approaches positive infinity, the numerator of the function approaches e^x, which grows without bound as x increases. The denominator also approaches infinity as x approaches positive infinity. Therefore, as x approaches positive infinity, the function approaches 1.
Thus, the horizontal asymptote is y = 1.
To find the vertical asymptote, we need to identify any values of x for which the denominator of the function is equal to zero. In this case, the denominator is 1+e^x. Setting this equal to zero and solving for x, we get:
1+e^x = 0
e^x = -1 (Note: e^x cannot be negative)
Since there are no values of x that satisfy this equation, there are no vertical asymptotes in this function.
To find the interval where the function is increasing, we need to identify the values of x where the derivative of the function is positive.
First, let's find the derivative of the function f(x):
f'(x) = (e^x)(1+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 find the interval where f(x) is increasing, we set f'(x) > 0 and solve for x:
e^x / (1+e^x)^2 > 0
Since e^x is always positive, we can ignore it in the inequality and focus on (1+e^x)^2 > 0.
(1+e^x)^2 > 0
Since (1+e^x)^2 is always positive, the inequality holds for all values of x. Therefore, the function is increasing for all intervals.
To find the inflection point, we need to determine the values of x where the concavity of the function changes. In other words, where the second derivative of the function changes sign.
First, let's find the second derivative of the function f(x):
f''(x) = [ e^x / (1+e^x)^2 ]'
Using the quotient rule, we can compute the derivative:
f''(x) = [(1+e^x)^2 - 2e^x(1+e^x)(e^x)] / (1+e^x)^4
= [(1+e^x)^2 - 2e^x(1+e^x)^2] / (1+e^x)^4
= [(1+e^x - 2e^x(1+e^x)] / (1+e^x)^3
= [(1+e^x - 2e^x - 2e^x)] / (1+e^x)^3
= [-e^x - e^(2x)] / (1+e^x)^3
Next, we set f''(x) = 0 and solve for x:
-e^x - e^(2x) = 0
Since e^x is always positive, we can divide both sides of the equation by -e^x:
1 + e^x = 0
This equation has no real solutions since e^x is always positive. Therefore, there are no inflection points in this function.
To find the interval where the function is concave up, we need to determine the values of x where the second derivative is positive.
Since the second derivative is always negative [f''(x) < 0], the function is always concave down, and there are no intervals where it is concave up.
To find the interval where the function is concave down, we can conclude that the entire domain of the function is concave down.