Let F X Be a Continuous Function With No Points of Inflection Which of the Following Must Bet Rue
Learning Objectives
- 4.5.1 Explain how the sign of the first derivative affects the shape of a function's graph.
- 4.5.2 State the first derivative test for critical points.
- 4.5.3 Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function's graph.
- 4.5.4 Explain the concavity test for a function over an open interval.
- 4.5.5 Explain the relationship between a function and its first and second derivatives.
- 4.5.6 State the second derivative test for local extrema.
Earlier in this chapter we stated that if a function has a local extremum at a point then must be a critical point of However, a function is not guaranteed to have a local extremum at a critical point. For example, has a critical point at since is zero at but does not have a local extremum at Using the results from the previous section, we are now able to determine whether a critical point of a function actually corresponds to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a graph by describing whether the graph of a function curves upward or curves downward.
The First Derivative Test
Corollary of the Mean Value Theorem showed that if the derivative of a function is positive over an interval then the function is increasing over On the other hand, if the derivative of the function is negative over an interval then the function is decreasing over as shown in the following figure.
A continuous function has a local maximum at point if and only if switches from increasing to decreasing at point Similarly, has a local minimum at if and only if switches from decreasing to increasing at If is a continuous function over an interval containing and differentiable over except possibly at the only way can switch from increasing to decreasing (or vice versa) at point is if changes sign as increases through If is differentiable at the only way that can change sign as increases through is if Therefore, for a function that is continuous over an interval containing and differentiable over except possibly at the only way can switch from increasing to decreasing (or vice versa) is if or is undefined. Consequently, to locate local extrema for a function we look for points in the domain of such that or is undefined. Recall that such points are called critical points of
Note that need not have local extrema at a critical point. The critical points are candidates for local extrema only. In Figure 4.31, we show that if a continuous function has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if has a local extremum at a critical point, then the sign of switches as increases through that point.
Using Figure 4.31, we summarize the main results regarding local extrema.
- If a continuous function has a local extremum, it must occur at a critical point
- The function has a local extremum at the critical point if and only if the derivative switches sign as increases through
- Therefore, to test whether a function has a local extremum at a critical point we must determine the sign of to the left and right of
This result is known as the first derivative test.
Theorem 4.9
First Derivative Test
Suppose that is a continuous function over an interval containing a critical point If is differentiable over except possibly at point then satisfies one of the following descriptions:
- If changes sign from positive when to negative when then is a local maximum of
- If changes sign from negative when to positive when then is a local minimum of
- If has the same sign for and then is neither a local maximum nor a local minimum of
We can summarize the first derivative test as a strategy for locating local extrema.
Problem-Solving Strategy
Problem-Solving Strategy: Using the First Derivative Test
Consider a function that is continuous over an interval
- Find all critical points of and divide the interval into smaller intervals using the critical points as endpoints.
- Analyze the sign of in each of the subintervals. If is continuous over a given subinterval (which is typically the case), then the sign of in that subinterval does not change and, therefore, can be determined by choosing an arbitrary test point in that subinterval and by evaluating the sign of at that test point. Use the sign analysis to determine whether is increasing or decreasing over that interval.
- Use First Derivative Test and the results of step to determine whether has a local maximum, a local minimum, or neither at each of the critical points.
Now let's look at how to use this strategy to locate all local extrema for particular functions.
Example 4.17
Using the First Derivative Test to Find Local Extrema
Use the first derivative test to find the location of all local extrema for Use a graphing utility to confirm your results.
Checkpoint 4.16
Use the first derivative test to locate all local extrema for
Example 4.18
Using the First Derivative Test
Use the first derivative test to find the location of all local extrema for Use a graphing utility to confirm your results.
Checkpoint 4.17
Use the first derivative test to find all local extrema for
Concavity and Points of Inflection
We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function.
Figure 4.34(a) shows a function with a graph that curves upward. As increases, the slope of the tangent line increases. Thus, since the derivative increases as increases, is an increasing function. We say this function is concave up. Figure 4.34(b) shows a function that curves downward. As increases, the slope of the tangent line decreases. Since the derivative decreases as increases, is a decreasing function. We say this function is concave down.
Definition
Let be a function that is differentiable over an open interval If is increasing over we say is concave up over If is decreasing over we say is concave down over
In general, without having the graph of a function how can we determine its concavity? By definition, a function is concave up if is increasing. From Corollary we know that if is a differentiable function, then is increasing if its derivative Therefore, a function that is twice differentiable is concave up when Similarly, a function is concave down if is decreasing. We know that a differentiable function is decreasing if its derivative Therefore, a twice-differentiable function is concave down when Applying this logic is known as the concavity test.
Theorem 4.10
Test for Concavity
Let be a function that is twice differentiable over an interval
- If for all then is concave up over
- If for all then is concave down over
We conclude that we can determine the concavity of a function by looking at the second derivative of In addition, we observe that a function can switch concavity (Figure 4.35). However, a continuous function can switch concavity only at a point if or is undefined. Consequently, to determine the intervals where a function is concave up and concave down, we look for those values of where or is undefined. When we have determined these points, we divide the domain of into smaller intervals and determine the sign of over each of these smaller intervals. If changes sign as we pass through a point then changes concavity. It is important to remember that a function may not change concavity at a point even if or is undefined. If, however, does change concavity at a point and is continuous at we say the point is an inflection point of
Definition
If is continuous at and changes concavity at the point is an inflection point of
Example 4.19
Testing for Concavity
For the function determine all intervals where is concave up and all intervals where is concave down. List all inflection points for Use a graphing utility to confirm your results.
Checkpoint 4.18
For find all intervals where is concave up and all intervals where is concave down.
We now summarize, in Table 4.1, the information that the first and second derivatives of a function provide about the graph of and illustrate this information in Figure 4.37.
Sign of | Sign of | Is increasing or decreasing? | Concavity |
---|---|---|---|
Positive | Positive | Increasing | Concave up |
Positive | Negative | Increasing | Concave down |
Negative | Positive | Decreasing | Concave up |
Negative | Negative | Decreasing | Concave down |
The Second Derivative Test
The first derivative test provides an analytical tool for finding local extrema, but the second derivative can also be used to locate extreme values. Using the second derivative can sometimes be a simpler method than using the first derivative.
We know that if a continuous function has local extrema, it must occur at a critical point. However, a function need not have local extrema at a critical point. Here we examine how the second derivative test can be used to determine whether a function has a local extremum at a critical point. Let be a twice-differentiable function such that and is continuous over an open interval containing Suppose Since is continuous over for all (Figure 4.38). Then, by Corollary is a decreasing function over Since we conclude that for all if and if Therefore, by the first derivative test, has a local maximum at On the other hand, suppose there exists a point such that but Since is continuous over an open interval containing then for all (Figure 4.38). Then, by Corollary is an increasing function over Since we conclude that for all if and if Therefore, by the first derivative test, has a local minimum at
Theorem 4.11
Second Derivative Test
Suppose is continuous over an interval containing
- If then has a local minimum at
- If then has a local maximum at
- If then the test is inconclusive.
Note that for case iii. when then may have a local maximum, local minimum, or neither at For example, the functions and all have critical points at In each case, the second derivative is zero at However, the function has a local minimum at whereas the function has a local maximum at and the function does not have a local extremum at
Let's now look at how to use the second derivative test to determine whether has a local maximum or local minimum at a critical point where
Example 4.20
Using the Second Derivative Test
Use the second derivative to find the location of all local extrema for
Checkpoint 4.19
Consider the function The points satisfy Use the second derivative test to determine whether has a local maximum or local minimum at those points.
We have now developed the tools we need to determine where a function is increasing and decreasing, as well as acquired an understanding of the basic shape of the graph. In the next section we discuss what happens to a function as At that point, we have enough tools to provide accurate graphs of a large variety of functions.
Section 4.5 Exercises
194 .
If is a critical point of when is there no local maximum or minimum at Explain.
195.
For the function is both an inflection point and a local maximum/minimum?
196 .
For the function is an inflection point?
197.
Is it possible for a point to be both an inflection point and a local extremum of a twice differentiable function?
198 .
Why do you need continuity for the first derivative test? Come up with an example.
199.
Explain whether a concave-down function has to cross for some value of
200 .
Explain whether a polynomial of degree can have an inflection point.
For the following exercises, analyze the graphs of then list all intervals where is increasing or decreasing.
202 .
204 .
For the following exercises, analyze the graphs of then list all intervals where
- is increasing and decreasing and
- the minima and maxima are located.
206 .
208 .
210 .
For the following exercises, analyze the graphs of then list all inflection points and intervals that are concave up and concave down.
212 .
214 .
For the following exercises, draw a graph that satisfies the given specifications for the domain The function does not have to be continuous or differentiable.
216 .
over over
217.
over over for all
218 .
over local maximum at local minima at
219.
There is a local maximum at local minimum at and the graph is neither concave up nor concave down.
220 .
There are local maxima at the function is concave up for all and the function remains positive for all
For the following exercises, determine
- intervals where is increasing or decreasing and
- local minima and maxima of
221.
over
222 .
For the following exercises, determine a. intervals where is concave up or concave down, and b. the inflection points of
223.
For the following exercises, determine
- intervals where is increasing or decreasing,
- local minima and maxima of
- intervals where is concave up and concave down, and
- the inflection points of
224 .
225.
226 .
227.
228 .
229.
230 .
For the following exercises, determine
- intervals where is increasing or decreasing,
- local minima and maxima of
- intervals where is concave up and concave down, and
- the inflection points of Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator.
231.
[T] over
232 .
[T] over
233.
[T] over
234 .
[T]
235.
[T]
236 .
[T] over
237.
over
238 .
239.
240 .
For the following exercises, interpret the sentences in terms of
241.
The population is growing more slowly. Here is the population.
242 .
A bike accelerates faster, but a car goes faster. Here Bike's position minus Car's position.
243.
The airplane lands smoothly. Here is the plane's altitude.
244 .
Stock prices are at their peak. Here is the stock price.
245.
The economy is picking up speed. Here is a measure of the economy, such as GDP.
For the following exercises, consider a third-degree polynomial which has the properties Determine whether the following statements are true or false. Justify your answer.
246 .
for some
247.
for some
248 .
There is no absolute maximum at
249.
If has three roots, then it has inflection point.
250 .
If has one inflection point, then it has three real roots.
Source: https://openstax.org/books/calculus-volume-1/pages/4-5-derivatives-and-the-shape-of-a-graph
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