Find the interval on which the curve y= integral x to 0 of 6/(1+2t+t^2) dt is concave upward
To determine the interval on which the curve y = ∫x to 0 of 6/(1 + 2t + t^2) dt is concave upward, we need to find the second derivative of the function.
Let's start by finding the first derivative:
dy/dx = d/dx (∫x to 0 (6/(1 + 2t + t^2)) dt)
Using the Fundamental Theorem of Calculus, we can differentiate under the integral sign:
dy/dx = 6/(1 + 2x + x^2)
Now, let's find the second derivative:
d²y/dx² = d/dx (dy/dx)
d²y/dx² = d/dx (6/(1 + 2x + x^2))
To differentiate this expression, we can use the Quotient Rule:
d²y/dx² = (6 * (-2)) / (1 + 2x + x^2)^2
Simplifying further, we get:
d²y/dx² = -12 / (1 + 2x + x^2)^2
Now, the curve y = ∫x to 0 of 6/(1 + 2t + t^2) dt will be concave upward where the second derivative is positive.
Therefore, we need to find the interval on which the expression -12 / (1 + 2x + x^2)^2 > 0.
To determine this interval, we need to find the roots of the denominator:
1 + 2x + x^2 = 0
This quadratic equation factors as:
(x + 1)(x + 1) = 0
Simplifying further, we get:
(x + 1)^2 = 0
This equation has a double root at x = -1. Therefore, the denominator does not change sign for any value of x.
Since the denominator is always positive, the second derivative -12 / (1 + 2x + x^2)^2 is always negative.
Therefore, there are no intervals on which the curve y = ∫x to 0 of 6/(1 + 2t + t^2) dt is concave upward.
To find the interval on which the curve is concave upward, we need to analyze the second derivative of the function.
1. Begin by finding the first derivative of the function. Since the integral is involved, we'll be using the Fundamental Theorem of Calculus to do this. Let's differentiate the function with respect to x.
First, rewrite the given function in terms of x:
y = ∫(x to 0) (6/(1+2t+t^2)) dt
To differentiate with respect to x, we need to use the Chain Rule:
dy/dx = d/dx ∫(x to 0) (6/(1+2t+t^2)) dt
Since the integral is dependent on the upper limit, we'll be using the Leibniz Rule for differentiating under the integral sign.
The Leibniz Rule states that if F(t, x) is continuous in a region containing a rectangle (a, b) × (c, d), x not in (a, b), f(t, x) is continuous in the region, and F(t, x) and its partial derivative ∂f/∂x are both continuous in the rectangle, then:
d/dx ∫(a to b) f(t, x) dt = ∫(a to b) ∂f/∂x(t, x) dt + f(b, x) - f(a, x).
Applying this rule to the given function, we have:
dy/dx = [∫(0 to x) ∂(6/(1+2t+t^2))/∂x dt] + (6/(1+2x+x^2)) - (6/(1+2(0)+0^2))
2. Now, let's compute the second derivative of y with respect to x. In order to be concave upwards, the second derivative should be positive. Differentiate dy/dx once again:
d^2y/dx^2 = d/dx[∫(0 to x) ∂(6/(1+2t+t^2))/∂x dt] + d/dx[6/(1+2x+x^2)]
3. Simplify the second derivative by taking the partial derivative and differentiating the second term:
∂(6/(1+2t+t^2))/∂x = 0
d/dx[6/(1+2x+x^2)] = -12(x+1)/(1+2x+x^2)^2
Therefore, the second derivative of y with respect to x is:
d^2y/dx^2 = -12(x+1)/(1+2x+x^2)^2
4. Now, we need to find the intervals where the second derivative is positive.
Since the denominator in the second derivative cannot be negative (it represents a squared term), we can focus on the numerator to determine the sign. The numerator, -12(x+1), will be positive when x < -1 and negative when x > -1.
Therefore, the curve is concave upward in the interval (-∞, -1) and concave downward in the interval (-1, ∞).
In conclusion, the interval on which the given curve y = ∫(x to 0) (6/(1+2t+t^2)) dt is concave upward is (-∞, -1).
y is concave up if y" > 0
By the 2nd FT of Calculus,
y' = -f(x) = -6/(x+1)^2
y" = 12(x+1)^3
so, where is y" positive?