Let f (x) = x^2 + 5x-14. 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 open intervals on which the function f(x) = x^2 + 5x - 14 is concave up or down, as well as the x-coordinates of any inflection points, we can follow these steps:
Step 1: Find the second derivative of f(x).
To determine the concavity of a function, we need to find its second derivative. The second derivative of f(x) is the derivative of the derivative of f(x).
The first derivative of f(x) is found by differentiating each term separately:
f'(x) = 2x + 5.
And the second derivative is found by differentiating the first derivative:
f''(x) = 2.
Step 2: Determine the concavity of f(x).
The concavity is determined by the sign of the second derivative. If f''(x) > 0, then the function is concave up on that interval. If f''(x) < 0, then the function is concave down on that interval.
Since the second derivative f''(x) = 2 is always positive, the function is concave up on the entire domain (all real numbers).
Step 3: Find the inflection points of f(x).
To find the inflection points, we need to identify the x-coordinates where the concavity changes. In other words, we need to find the values of x for which the second derivative is equal to zero or undefined.
Since the second derivative f''(x) is a constant 2 and does not change sign, there are no inflection points in this case.
Therefore, the answers to the given questions are:
1. f is concave up on the intervals: (-∞, ∞) (the entire real number line).
2. f is concave down on the intervals: None (since f is concave up on the entire domain).
3. The function f does not have any inflection points.
f=x^2+5x-14
f '=2x+5
f ''=2
since f'' > 0, always concave up (I mean, come on, it's a parabola!)
so, no inflections
this should have been a no-brainer