Find the largest open intervals where the function is concave upward.
f(x)= x/x^2+1
I think you meant:
f(x) = x/(x^2 + 1)
(or else why not just reduce it to
f(x) = 1/x + 1 ? )
for concave upwards, f ''(x) > 0
f'(x) = (1 - x^2)/(x^2 + 1)^2 using the quotient rule
and again
f''(x) = 2x(x^2-3)/(x^2+1)^3
so when is f''(x) positive ?
well, the denominator is always positive,
so we just have to look at the top
let's find the points of inflection:
2x = 0 ---> x = 0, then y = 0
x^2 - 3 = 0
x = ± √3 , then y = ±√3/4
look at the original graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%2F%28x%5E2+%2B+1%29
opens up for x >√3/4
and -√3/4 < x < 0
the first of these being the largest open interval where it opens upwards.
Strange wording of the question.
To find the intervals where the function f(x) = x/(x^2 + 1) is concave upward, we need to determine where the second derivative is positive.
Step 1: Find the first derivative of f(x)
To find the first derivative, we will use the Quotient Rule. The Quotient Rule states that if we have a function u/v, where u and v are differentiable functions, then the derivative of u/v is given by:
f'(x) = (v * u' - u * v') / v^2
In our case, u = x and v = x^2 + 1. Taking the derivatives, we get:
u' = 1
v' = 2x
Applying the Quotient Rule, we get:
f'(x) = (x^2 + 1 - 2x^2) / (x^2 + 1)^2
f'(x) = (1 - x^2) / (x^2 + 1)^2
Step 2: Find the second derivative of f(x)
To find the second derivative, we differentiate f'(x):
f''(x) = [(1 - x^2)'(x^2 + 1)^2 - (1 - x^2)'(2(x^2 + 1)(2x))] / (x^2 + 1)^4
f''(x) = [(0 - 2x)(x^2 + 1)^2 - (1 - x^2)(4x^3 + 4x)] / (x^2 + 1)^4
f''(x) = [(2x^3 + 2x) - (4x^3 + 4x - 4x^5 - 4x)] / (x^2 + 1)^4
f''(x) = (2x^3 - 4x^5) / (x^2 + 1)^4
Step 3: Determine where the second derivative is positive
To find the intervals where f(x) is concave upward, we need to determine where the second derivative f''(x) is positive.
Setting the second derivative f''(x) greater than 0:
(2x^3 - 4x^5) / (x^2 + 1)^4 > 0
Next, we need to find the critical points where the numerator and denominator equal zero.
For f''(x) = 0,
2x^3 - 4x^5 = 0
2x(x^2 - 2x^4) = 0
Setting each factor equal to zero, we get:
2x = 0, which gives us x = 0 (a critical point)
x^2 - 2x^4 = 0
x^2(1 - 2x^2) = 0
Setting each factor equal to zero, we get:
x^2 = 0, which gives us x = 0 (a repeated critical point)
1 - 2x^2 = 0
-2x^2 = -1
x^2 = 1/2
x = ±√(1/2)
Now, we can create a sign chart to determine the intervals where f''(x) is positive or negative:
x | 0 | √(1/2) | -√(1/2) |
---------------------------------------
f''(x) | 0 | + | - | + |
From the sign chart, we can see that f''(x) is positive for x > √(1/2) and x < -√(1/2).
Therefore, the largest open intervals where the function f(x) = x/(x^2 + 1) is concave upward are (-∞, -√(1/2)) and (√(1/2), ∞).
To find the largest open intervals where the function is concave upward, we need to determine where the second derivative of the function is positive.
Let's start by finding the first and second derivatives of the function f(x):
f(x) = x/(x^2 + 1)
First derivative:
f'(x) = [(1)(x^2 + 1) - (x)(2x)] / (x^2 + 1)^2
= (x^2 + 1 - 2x^2) / (x^2 + 1)^2
= (1 - x^2) / (x^2 + 1)^2
Second derivative:
f''(x) = [(1 - x^2)(2(x^2 + 1)(2x)) - (1 - x^2)(2x)(2x)] / (x^2 + 1)^4
= [(1 - x^2)(2(x^2 + 1)(2x) - 2x(2x))] / (x^2 + 1)^4
= [(1 - x^2)(4x^3 + 4x - 4x^2 - 4x)] / (x^2 + 1)^4
= [4x(x^3 + 1 - x^2 - x)] / (x^2 + 1)^4
= [4x(x^3 - x^2 - x + 1)] / (x^2 + 1)^4
Now, we set the second derivative equal to zero and solve for x to find the possible points of inflection:
4x(x^3 - x^2 - x + 1) = 0
The solutions to this equation will give us the potential critical points for concavity.
By factoring out x = 0, we get:
x = 0
Now let's analyze the intervals separately:
For x < 0, the second derivative term has the same sign as x(x^3 - x^2 - x + 1). In this case, the sign of the function is determined by the sign of the expression inside the brackets. This can be determined by choosing a point within each interval (excluding x = 0) and evaluating it.
If we take x = -1, then the expression becomes:
-(-1)(-1^3 - (-1)^2 - (-1) + 1) = 1(1 - 1 + 1 + 1) = 1(2) = 2
Since the expression is positive, the concavity is upward for x < 0.
For x > 0, the same analysis can be applied. Let's choose x = 1:
1(1^3 - 1^2 - 1 + 1) = 1(1 - 1 - 1 + 1) = 1(-1) = -1
Since the expression is negative, the concavity is downward for x > 0.
Therefore, we can conclude that the largest open interval where the function f(x) = x/(x^2 + 1) is concave upward is (-∞, 0).