Let F(x)=int from 0 to x of (t-3)/(t^2+7) dt for -infinity <x< infinity.
a. Find the value of x where F attains its minimum value.
b. find the intervals over which F is only increasing or only decreasing.
c. Find open intervals over which F is only concave up or only concave down.
To find the value of x where F attains its minimum value, we need to find the critical points of the function F(x).
a. To find the critical points, we first find the derivative of F(x):
F'(x) = d/dx ∫[0 to x] (t-3)/(t^2+7) dt
To differentiate this integral, we will use the Fundamental Theorem of Calculus:
F'(x) = (x-3)/(x^2+7)
To find the critical points, we set F'(x) equal to zero and solve for x:
(x-3)/(x^2+7) = 0
Since the numerator can only be equal to zero, we have:
x - 3 = 0
x = 3
Therefore, the value of x where F attains its minimum value is x = 3.
b. To find the intervals over which F is only increasing or only decreasing, we need to analyze the sign of the derivative F'(x).
From part a, we know that F'(x) = (x-3)/(x^2+7).
To determine the sign of F'(x), we need to examine when the numerator and denominator are positive or negative.
When (x-3) > 0 and (x^2+7) > 0:
x > 3 and x^2 > -7
Since x^2 is always positive, this condition is satisfied for any x > 3.
When (x-3) < 0 and (x^2+7) > 0:
x < 3 and x^2 > -7
Again, x^2 is always positive, so this condition is satisfied for any x < 3.
Therefore, F is only increasing for x > 3 and only decreasing for x < 3.
c. To find the open intervals over which F is only concave up or only concave down, we need to analyze the sign of the second derivative F''(x).
To find the second derivative, we differentiate F'(x):
F''(x) = d/dx (x-3)/(x^2+7)
F''(x) = [1*(x^2+7) - (x-3)*(2x)] / (x^2+7)^2
F''(x) = (x^2 + 7 - 2x^2 + 6x) / (x^2+7)^2
F''(x) = (-x^2 + 6x + 7) / (x^2+7)^2
To determine the sign of F''(x), we analyze when the numerator and denominator are positive or negative.
For F''(x) to be positive, the numerator, -x^2 + 6x + 7, should be positive. We can solve this:
-x^2 + 6x + 7 > 0
(x - 7)(-x + 1) > 0
Solving this quadratic inequality, we find that the intervals where F''(x) is positive are:
x < 1 or x > 7
For F''(x) to be negative, the numerator, -x^2 + 6x + 7, should be negative.
We can determine this by checking the intervals between the roots of the quadratic equation:
1 < x < 7
Therefore, the open intervals over which F is only concave up are:
(-∞, 1) U (7, +∞)
The open interval over which F is only concave down is:
(1, 7)