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# Sketch the graph of a function that satisfies all of the given conditions$f'(x) > 0$ for all $x \not= 1$, vertical asymptote $x = 1$,$f"(x) > 0$ if $x < 1$ or $x > 3$, $f"(x) < 0$ if $1 < x < 3$

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So we're given a bunch of conditions that we need to fulfill for our graph and the first one is that are derivative is greater than zero for all X. That is not equal to one. So our function is increasing for all X, not equal to one. We have a vertical as um. Tota at X is equal to one. And then our second derivative is greater than zero when X is less than one or when X is greater than three. And our second derivative is less than zero when X is between one and 3. So we're concave up from negative infinity to one and from three to infinity. And we are concave down from 1 to 3. So if we do this, we know that we're increasing for all X. And we have this vertical ascent to it at one and we are concave up before one. So we need a curve that is always increasing in concave up. It has this assume tote. So it's going to look something like this where we're concave up were increasing and we have this assam tote that we're increasing towards and we're never going to cross. And then if we look at values of X that are greater than one, but less than three, we know that we're concave down. So we're concave down but we also know that we are increasing since we're increasing everywhere, that X is not equal to one. So for concave down until X is equal to three, it would look something like this and then We'd have this inflection point at X is equal to three right here where we go from being concave down to concave up however, were increasing for the entirety of this part of our graph. So we're not it's not a local maximum minimum, it's an inflection point. And so this would be a good sketch of the graph given our conditions where we're increasing everywhere, where concave up before this ascent to X is equal to one, and then we're concave down from 1-3, and we're concave up again from three to infinity.

Oregon State University

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