Designer functions Sketch the graph of a function f that ...

Question

Designer functions Sketch the graph of a function f that is continuous on (-∞,∞) and satisfies the following sets of conditions.

f"(x) > 0 on (-∞,-2); f"(-2) = 0; f'(1) = 0; f"(2) = 0; f'(3) = 0; f"(x) > 0 on (4,∞)

Accepted Answer

Below there today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Which of the following graphs of FXX satisfies the following conditions? F of X is greater than 0 negative infinity -1, F-1 is equal to 0. F is equal to F 2 is equal to 0. F of 3 is equal to 0. F 4 is equal to 0. F of X is greater than 0 on parentheses 5 infinity. Awesome, so it appears for this particular problem we're trying to determine which of the following graphs and it appears we're given 4 separate graphs. Which of these 4 graphs satisfies all of these conditions that are set to us by the prom itself. So looking briefly at all of our different graphs that are given to us by the problem itself, we can see that graphs B and D both point upwards, whereas B crosses through the origin 0.00 and graph D does not. And graphs A and C both point downwards. When I say they both point downwards for both all of our multiple choice answers, I'm saying that the curve where it starts and ends both point downwards. So the curve for A and C both point downwards. However, for the curve for a crosses through the origin point, and the curve for graph C does not cross through the origin, but both graphs A and C, both of the. Curves point, so the both of the ends for the curve point downwards, whereas the curves for B and D both point upwards. So the curves begins and ends in the upwards direction. And once again B crosses through the origin, whereas graph D does not. OK, so now that we have a good visualization of what's going on for this particular prompt, our first step that we need to take in order to solve for this problem is we need to identify the given condition for the function F of X and what they apply about its graph. So we need to determine what this implies about the graph at large. So we need to recall and note that for F of X is greater than 0 on negative infinity 1 in parentheses, where parentheses means it's an open interval, whereas if it was square brackets, it would be a closed interval. And this 5 com infinity, so once again. This F of X is greater than 0 on negative infinity-1 parentheses and 5 infinity in parentheses. This indicates that the graph is concave up in these intervals. So let's make a note of this CU we'll call this, this CU denotes concave up. Fantastic. And F of -1 is equal to 0 and F of 3 equals to 0. This suggests that there's an inflection point at X equals -1, and X equals 3. Where there is where the concavity of FFX changes. So once again, this F of -1 equals 0 and F of 3 equals 0 suggests that there will be an inflection point at -1 and 3, so X equals -1 and X equals 3, and this is where the concavity of F of X changes. Let us also note that F of 0 and F of 2. is equal to 0 and F of 4 equals 0 indicates that there will be a horizontal tangents or critical points at X equals 0, X equals 2, and X equals 4. So once again, to quickly recap here, So where F of 0 is equal to F 2 is equal to 0 and F of 4 is equal to 0, this will indicate a horizontal tangents are visible, or there will be critical points that are located at X equals 0, X equals 2, and X equals 4, where the slope of this line tangent to the graph will also be 0. And these could be local maxima, minima, or inflection points. So now, with that said, let us analyze each of our options. So for option A, we can see that on the interval negative infinity 1 in parentheses, the function is concave down like we were discussing earlier, both the starting point of the curve and the ending point of the curve points downwards. So from these given conditions of F of X on the interval parentheses negative infinity.1, the function should be concave up. So therefore, this does not satisfy the given conditions. So our final answer cannot be multiple choice answer A. For option B, on the intervals negative infinity -1, the function is concave up, as we said, the curve the curve starts upwards and ends in the upwards direction. And on the interval 5 infinity, the function is initially concave down but switches to concave up, and from the given conditions on the interval parentheses 5 infinity, the function should be concave only. So therefore, this does not satisfy the given condition, so B cannot be our final answer. Multiple choice answer C, as we can see on the interval, negative infinity 1 in parenthesis. The function is concave down, as you can see the curve starts in the downwards position and ends in the downwards position. And from the given conditions F of X on the intervals negative infinity com-1 should be concave up. So therefore this does not satisfy the given conditions. Multiple choice answer letter D on the intervals negative infinity 1 in parentheses and. 5 infinity in parentheses. The function is concave up. There are also inflection points at -1 and 3. So at X equals -1 and X equals 3. So let's make a note of that. So X equals -1 and X equals 3, and there is local maxima at X equals 0.2.4, so X equals 0 and X equals 2 and X equals 4. So therefore, without a doubt, this satisfies our given condition. So our final answer has to be multiple choice answer letter D, and that's it. We've officially solved for this problem. So in order to get a better view of this graph, I blew it up once again. And as you can see, Our multiple choice answer letter D is the only graph that satisfies all the conditions that are set to us by the problem itself. So that means without a doubt, our final answer has to be multiple choice answer, letter D. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Designer functions Sketch the graph of a function f that is continuous on (-∞,∞) and satisfies the following sets of conditions.

f"(x) > 0 on (-∞,-2); f"(-2) = 0; f'(1) = 0; f"(2) = 0; f'(3) = 0; f"(x) > 0 on (4,∞)

Key Concepts

Second Derivative Test

Critical Points

Continuity and Differentiability

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