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, for all x in the domain of f'; f'(-2) and f'(1) do not exist; f"(0) = 0

Accepted Answer

Hello, for this video, we are going to be selecting the graph that matches the given information. We are asked to find which of the following graphs of G of X that satisfies the following conditions. We are told that the derivative of G of X is less than 0 for all X in the domains of G of X. We are also told that G -1 and G-7 do not exist, and G of 3 is equal to 0. Let's go ahead and start with the last condition. G 3 equal to 0. Now, what does it mean for the second derivative to equal to 0? Well, when the second derivative equals to 0, that means that there is an inflection point at this value of X. When there is an inflection point, that means that the concavity of the graph is going to switch for the original graph G of X. That can look like concave up switching to concave down, or it can even look as concave down, switching to concave up. So let's go ahead and take a look at the graphs at the value of 3, which looks like the concavity is switching. If we look at option A, notice that the that this graph switches from concave up to concave down at the value of 3. That means option A is a potential solution. We can witness the same behavior for the second option B. We switch from concave down to concave up at the value X is equal to 3. Now for options C and D, this is where the behaviors get a little tricky. The entirety of the behavior from -1. 26 is all concave up for option C. That does not match this behavior of G3 equal to 0, so option C will not be the solution. And the same can be said for option D. We do not have a switch of concavity when we pass through the value of 3. The graph stays concave down, so option D will not be a solution. So now we have to determine if option A or option B is the correct solution. Well, let's take a look at the 2nd given condition. We are told that G of -1 and G of 7. Do not exist now. What does it mean for the derivative to not exist at a value? Well, there are two conditions where a graph is not differentiable. We have the first condition where we have a jump discontinuity in the graph. Wherever there is a jumped discontinuity, the derivative does not exist. And the other condition is when we have some type of cusp. If we ever have a cusp on the point of a graph, the derivative does not exist at the point of the cusp. If we take a look at option A, option A has a point, a cusp at -1, and has a cusp at 7. And the same can be true for option B. There is a cusp at -1 and there is a cusp at 7. So both option A and B satisfy the second condition. Now, what about the very first condition? G of X is less than 0 for all X within the domain. Now what this means is that the derivative will always be decreasing. For or be negative for the graph of G of X. What this means is that the original graph G of X will be decreasing for the entirety of its graph, and the graph that satisfies this condition is option A. Starting from negative infinity and moving to positive infinity, the graph is slowly decreasing over time, whereas in option B, the opposite is witnessed. We see a decrease going to -1, then we increase all the way to 7, then decrease to infinity after 7, which means that option A is going to be our solution because it satisfies all three conditions given to us in this problem. So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Designer functions Sketch the graph of a function f that is continuous on (-∞,∞) and satisfies the following sets of conditions.f'(x) > 0, for all x in the domain of f'; f'(-2) and f'(1) do not exist; f"(0) = 0

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

f'(x) > 0, for all x in the domain of f'; f'(-2) and f'(1) do not exist; f"(0) = 0