How do you find points of inflection and determine the intervals of concavity given #y=x^2-x-1#?

1 Answer
Oct 4, 2017

No inflection points exist, function is convex (aka concave up) for all x.


If we use a graph, the concavity change and inflection points will be self evident. A function is concave up (aka convex) on the interval (a,b) if a line segment passing through both a and b lies above the function ; the function is concave downwards when the function instead lies above the segment.

graph{x^2-x-1 [-10, 10, -5, 5]}

Here we see the function is always concave upwards. Since this is true, there is no inflection point; for an inflection point to exist, the concavity must change. If there is no change in concavity, there is no inflection point.

We can prove this in a non graphical manner as well. Our function is #f (x) = x^2-x-1 -> f'(x) = 2x-1 -> f''(x) = 2.# It is necessary for #f''(x_o)=0# in order for #x_0# to be an inflection point, though it should be noted that this is a necessary condition, not a sufficient one. Since the second derivative is constant and greater than 0, it is convex (aka concave up) and never changes its concavity; it is concave up on #(-oo,oo)#.