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# Sketch the graph of a function that satisfies all of the given conditions(a) $f'(x) > 0$ and $f"(x) < 0$ for all $x$(b) $f'(x) < 0$ and $f"(x) > 0$ for all $x$

## (a) $f^{\prime}(x)>0$ and $f^{\prime \prime}(x)<0$ for all $x$The function must be always increasing (since the first derivative is alwayspositive) and concave downward (since the second derivative is alwaysnegative).(b) $f^{\prime}(x)<0$ and $f^{\prime \prime}(x)>0$ for all $x$The function must be always decreasing (since the first derivative is alwaysnegative) and concave upward (since the second derivative is alwayspositive

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So we're given these two conditions that the derivative of a function F. Of X is greater than zero and the second derivative of our function F. Of X is less than zero for all X. And so what the derivative being greater than zero means is that we're increasing for the entirety of this function. And what the second derivative being less than zero means is that we are concave down for the entirety. So this first derivative, it always tells us if we're increasing or decreasing, second derivative tells us about con cavity. So if we're going to draw a sketch of this graph, we want a concave down graph that is always increasing. So we're gonna want a graph that looks kind of like this where we're still concave down where an upside down U. Shaped type of graph. But we never actually um go decreasing. So we never actually, it's not like a problem where we would come back and actually have this upside down to you. It's just the curvature is like an upside down U. So this would be a potential sketch of a graph for a function that is always increasing in concave down. And for part B we are told that are derivative of prime of X is less than zero, and our second derivative is greater than zero for all X. And so we're going to be decreasing for the entirety of our function and concave up. So we want to sketch a graph that is both decreasing in concave up. So what that would look like is something like this where you start like this and concave up. So it's like a curvature for a regular you and we're always decreasing. This isn't going to be a problem, it's going to keep decreasing for the entirety of the function and it's going to always be concave up.

Oregon State University

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