Let f (x) =7x-5/x+3
. Find the open intervals
on which f is concave up (down). Then determine the xcoordinates
of all inflection points of f .
1. f is concave up on the intervals
2. f is concave down on the intervals
3. The inflection points occur at x =
To find the intervals on which f is concave up or down, we need to determine the second derivative of the function f(x).
1. Find the first derivative:
f'(x) = [7(x + 3) - (7x - 5)] / (x + 3)^2
= 7(x + 3 - x + 5) / (x + 3)^2
= 12 / (x + 3)^2
2. Find the second derivative:
f''(x) = d/dx (12 / (x + 3)^2)
= (-12(2)(x + 3)) / (x + 3)^4
= -24 / (x + 3)^3
To determine the concavity, we need to evaluate the sign of f''(x).
If f''(x) > 0, the function is concave up on that interval.
If f''(x) < 0, the function is concave down on that interval.
To find the intervals, we need to find where f''(x) = 0, which are the potential inflection points.
3. Set f''(x) = 0 and solve for x:
-24 / (x + 3)^3 = 0
Since a fraction is equal to zero only when the numerator is zero, we have:
-24 = 0
This is not possible, so f''(x) does not equal zero, and we have no inflection points.
Therefore:
1. f is concave up on the intervals: (-∞, -3) and (-3, ∞)
2. f is concave down on the interval: (-3, ∞)
3. There are no inflection points for the function f.