Like

Report

In each part state the $ x $-coordinates of the inflection points of $ f $. Give reasons for your answers.

(a) The curve is the graph of $ f $.

(b) The curve is the graph of $ f' $.

(c) The curve is the graph of $ f" $.

(a) There is an IP at $x=3$ because the graph of $f$ changes from $\mathrm{CD}$ to $\mathrm{CU}$ there. There is an $\mathrm{IP}$ at $x=5$ because the graph

of $f$ changes from $\mathrm{CU}$ to $\mathrm{CD}$ there.

(b) There is an IP at $x=2$ and at $x=6$ because $f^{\prime}(x)$ has a maximum value there, and so $f^{\prime \prime \prime}(x)$ changes from positive to

negative there. There is an IP at $x=4$ because $f^{\prime}(x)$ has a minimum value there and so $f^{\prime \prime}(x)$ changes from negative to

positive there.

(c) There is an inflection point at $x=1$ because $f^{\prime \prime}(x)$ changes from negative to positive there, and so the graph of $f$ changes

from concave downward to concave upward. There is an inflection point at $x=7$ because $f^{\prime \prime}(x)$ changes from positive to

negative there, and so the graph of $f$ changes from concave upward to concave downward.

You must be signed in to discuss.

So in this problem we are given this graph and we're told this graph is either going to be our function F of X, the first derivative of our function F. Of X, or the second derivative or function F of X. And we want to find the inflection points given this graph is one of those functions. So in part they were going to be assuming that this is the graph of our function F. Of X. And we want to find the inflection points. So the inflection points of our function F of X are going to be at the point where we change in con cavity. So when we go from concave up to concave down or from concave down to concave up and it might be a little bit harder to figure that out with my sketch, but it was a little bit more um clear in the actual graph of this that the inflection point would be At this point here, when X is equal to three, and at this point here, when X is equal to four, or siren X is equal to five. And then those are the only two inflection points. So the reason that I know this is because we have a concave down graph that then changes to concave up here and now we're concave up and then we change again to concave down at this point here. So since we're changing incan cavity, that means that these are inflection points. So And say the inflection points, RIPS are at X is equal to three And X is equal to five. And for part B, we're going to figure out where the inflection points are given that this is the derivative of our function F of X. So if we look at this graph again, the inflection points or how to find an inflection point, given that this is the derivative of our function F of X is going to be wherever our function has a maximum or minimum value, that's going to be where we have inflection points for our function F. X. And the reason for that is because we're going to be having zeros for the derivative of our derivative of F of X. Or for a second derivative at these maximum minimum values. And we're also going to be going from positive to negative or negative to positive. In terms of our second derivative at these maximum minimum values, we're not going to be going from positive to then hitting zero to then positive again for our second derivative, since these are maximum minimum values. So the maximum minimum values of our function occur at x is equal to two, at X is equal to four and at X is equal to six. So our inflection points Are at X is equal to two. X is equal to four and acts as equal to sex. And for part C. We're going to assume that this is the second derivative of our function F. X. And to find the inflection points, given that this is the second derivative of our function F of X. Those are going to be at the zeros of our function F double prime of X. But they also have to go from negative to positive Are positive and negative. We have to have a sign change after this. zero for our for it to be an inflection point. So that only occurs here at X is equal to one and at X is equal to seven since it X is equal to four. At this point it is a zero. But we go from positive values to then positive values again. So we're not actually changing con cavity. We're not going from concave up to concave down. So the two inflection points, given that this is the second derivative of a function are going to be, at X is equal to one, and at X is equal to seven.