The blue colour indicates a region where the slope of the tangent decreases. That is, in this region the rate at which the cost function increases, decreases. The red colour indicates a region where the slope of the tangent increases, i.
By our previous definitions, the blue area is concave downward and the red area is concave upwards. The green point is the point at which the rate of change of the slope changes from decreasing to increasing. It is also the point at which the concavity of the function changes from downward to upward.
This point is called a point of inflection POI. In general, note that regardless of the sign of the slope positive, negative or zero , the slopes of the tangent are decreasing as we move from left to right when the graph is concave down and increasing from left to right when it is concave up. Look at the applet below. Note that the function is shown in the top line, the function's graph on the left and it's derivative's graph on the right.
Use the slider to change the value of x. Convince yourself that the graph of the given function f is concave up where the derivative f ' the slope of the tangent is an increasing function, and concave down where the derivative f ' is a decreasing function.
Examine the fourteen examples provided in the scroll bar on the top of the applet or enter your own function in the space below the graphs and press new function to display the graphs. Possible Answers:. Correct answer:.
Explanation : To find when a function is concave, you must first take the 2nd derivative, then set it equal to 0, and then find between which zero values the function is negative. First, find the 2nd derivative: Set equal to 0 and solve: Now test values on all sides of these to find when the function is negative, and therefore decreasing. Report an Error. Possible Answers: No, is positive on the interval.
Correct answer: Yes, is negative on the interval. So, for all So on the interval -5,-4 f x is concave down because f'' x is negative. Possible Answers: Concave down, because is negative on the given interval. Correct answer: Concave up, because is positive on the given interval.
Explanation : To test concavity, we need to perform the second derivative test. Begin as follows: Next, we need to evaluate h" t on the interval [5,7] So, our h" t is positive on the interval, and therefore h t is concave up.
Explanation : The intervals where a function is concave up or down is found by taking second derivative of the function. Use the power rule which states: Now, set equal to to find the point s of infleciton. Explanation : To find which interval is concave down, find the second derivative of the function. Now to find which interval is concave down choose any value in each of the regions , and and plug in those values into to see which will give a negative answer, meaning concave down, or a positive answer, meaning concave up.
Possible Answers: It is never concave down. Explanation : The derivative of is The derivative of this is This is the second derivative. A function is concave down if its second derivative is less than 0. Explanation : To find the concavity of a graph, the double derivative of the graph equation has to be taken.
We also must remember that the derivative of an constant is 0. After taking the first derivative of the equation using the power rule, we obtain.
The double derivative of the equation we are given comes out to. Determine the intervals on which the following function is concave down :. Explanation : To find the invervals where a function is concave down, you must find the intervals on which the second derivative of the function is negative. The second derivative of the function is equal to. Both derivatives were found using the power rule. How many infelction points does the function have on the interval? Possible Answers: One. Correct answer: Three.
Explanation : Points of inflection occur where there second derivative of a function are equal to zero. Taking the first and second derivative of the function, we find: To find the points of inflection, we find the values of x that satisfy the condition.
Which occurs at Within the defined interval [-5, 5], there are three values:. Copyright Notice. View Calculus Tutors. Avianna Certified Tutor. Lei Certified Tutor. Byron Certified Tutor. University of Miami, Master of Science, Mathematics.
Report an issue with this question If you've found an issue with this question, please let us know. Do not fill in this field. Louis, MO Or fill out the form below:. A function can be concave up and either increasing or decreasing.
Similarly, a function can be concave down and either increasing or decreasing. Here is the mathematical definition of concavity. To show that the graphs above do in fact have concavity claimed above here is the graph again blown up a little to make things clearer.
So, as you can see, in the two upper graphs all of the tangent lines sketched in are all below the graph of the function and these are concave up. In the lower two graphs all the tangent lines are above the graph of the function and these are concave down. This is important to note because students often mix these two up and use information about one to get information about the other. Now that we have all the concavity definitions out of the way we need to bring the second derivative into the mix.
We did after all start off this section saying we were going to be using the second derivative to get information about the graph.
The following fact relates the second derivative of a function to its concavity. So, what this fact tells us is that the inflection points will be all the points were the second derivative changes sign. We saw in the previous chapter that a function may change signs if it is either zero or does not exist. It is simply a fact that applies to all functions regardless of whether they are derivatives or not. We will only know that it is an inflection point once we determine the concavity on both sides of it.
It will only be an inflection point if the concavity is different on both sides of the point. If you think about it this process is almost identical to the process we use to identify the intervals of increasing and decreasing.
This only difference is that we will be using the second derivative instead of the first derivative. The first thing that we need to do is identify the possible inflection points. The second derivative in this case is a polynomial and so will exist everywhere. It will be zero at the following points.
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