How to calculate the lowest repeatable time

Calculate extreme values

In this chapter you will learn how to calculate the extreme values ​​of a function. From a graphical point of view, these are high points or low points.

There are basically two different approaches to calculating the extreme values ​​of a function. The difference between the two methods is the use of the second derivative. With one method you have to calculate the second derivative, with others you can save the second derivative.

Calculate extreme values ​​- with 2nd derivative!

In school you usually learn to calculate extreme values ​​with the help of the second derivative.

In this context, you should know the following definitions:

A High point is present if the following applies:

\ (f '(x_0) = 0 \ qquad \ text {and} \ qquad f' '(x_0) <0 \)

A Low point is present if the following applies:

\ (f '(x_0) = 0 \ qquad \ text {and} \ qquad f' '(x_0)> 0 \)

What might look a bit cryptic at first glance is actually quite simple:

  1. Calculate the first derivative
  2. Calculate zeros of the first derivative
  3. Calculate the second derivative
  4. Insert the zeros of the first derivative into the second derivative
    \ (\ rightarrow \) is the second derivative then less than zero, it is one High point
    \ (\ rightarrow \) is the second derivative greater than zero, it is one Low point
  5. Calculate the y-coordinates of the high points / low points

example 1

The function \ (f (x) = x ^ 2 \) is to be examined for extreme values.

1.) Calculate the first derivative

\ (f '(x) = 2x \)

2.) Calculate zeros of the first derivative

Approach: \ (f '(x) = 0 \)

\ (f '(x) = 2x = 0 \ qquad \ rightarrow \ qquad x = 0 \)

3.) Calculate the second derivative

\ (f '' (x) = 2 \)

4.) Insert the zeros of the first derivative into the second derivative

Since there is no x in the second derivative, we are already done!

The second derivative is always greater than zero: \ (f '' (x) = 2> 0 \).

... for this reason there is a low point at the point \ (x = 0 \).

5.) Calculate the y-coordinate of the low point

Our task is to calculate a high POINT or a low POINT. A point always determines from two coordinates, which is why you should not forget to calculate the y-coordinate!

To do this, we insert the already known x-value of the low point into the original function \ (f (x) \):

\ (y = f (0) = 0 ^ 2 = 0 \)

Summary

The function has a low point at the point (0 | 0).

The function \ (f (x) = x ^ 2 \) is drawn in the coordinate system. In addition, the extreme value (= lowest point) of the function is marked in red.

Example 2

The function \ (f (x) = \ frac {2} {3} x ^ 3 + 3x ^ 2 + 4x \) is to be examined for extreme values.

1.) Calculate the first derivative

\ (f '(x) = 2x ^ 2 + 6x + 4 \)

2.) Calculate zeros of the first derivative

Approach: \ (f '(x) = 0 \)

\ (f '(x) = 2x ^ 2 + 6x + 4 = 0 \)

It's a quadratic equation that we solve using the midnight formula. Alternatively, one could use the pq formula or Vieta's theorem, for example.

\ [x_ {1,2} = \ frac {-b \ pm \ sqrt {b ^ 2 - 4ac}} {2a} = \ frac {-6 \ pm \ sqrt {6 ^ 2 - 4 \ cdot 2 \ cdot 4}} {2 \ cdot 2} = \ frac {-6 \ pm 2} {4} \]

\ [x_ {1} = \ frac {-6 - 2} {4} = -2 \]

\ [x_ {2} = \ frac {-6 + 2} {4} = -1 \]

3.) Calculate the second derivative

\ (f '' (x) = 4x + 6 \)

4.) Insert the zeros of the first derivative into the second derivative

\ (f '' (- 2) = 4 \ cdot (-2) + 6 = -2 <0 \ qquad \ text {at the point \ (x = -2 \) is a high point} \)

\ (f '' (- 1) = 4 \ cdot (-1) + 6 = 2> 0 \ qquad \ text {at the point \ (x = -1 \) there is a low point} \)

5.) Calculate the y-coordinate of the high point / low point

It is our task to calculate a high POINT or a low POINT. A point always determines from two coordinates, which is why you should not forget to calculate the y-coordinate!

To do this, we insert the already known x-value of the high point / low point into the original function \ (f (x) \):

\ (y = f (-2) = \ frac {2} {3} \ cdot (-2) ^ 3 + 3 \ cdot (-2) ^ 2 + 4 \ cdot (-2) = - \ frac {4 } {3} \)

\ (y = f (-1) = \ frac {2} {3} \ cdot (-1) ^ 3 + 3 \ cdot (-1) ^ 2 + 4 \ cdot (-1) = - \ frac {5 } {3} \)

Summary

The function has a high point at \ (\ left (-2 | - \ frac {4} {3} \ right) \).

The function has a low point at \ (\ left (-1 | - \ frac {5} {3} \ right) \).

The function \ (f (x) = \ frac {2} {3} x ^ 3 + 3x ^ 2 + 4x \) is drawn in the coordinate system. In addition, the extreme values ​​of the function are marked in red.

Calculate extreme values ​​- without 2nd derivative!

Another possibility to calculate the extreme values ​​of a function is based on the investigation of the monotony behavior.

You should be familiar with the following definitions:

The function \ (f \) is strictly monotonically increasingif \ (f '(x)> 0 \) holds.

The function \ (f \) is strictly monotonically falling, if \ (f '(x) <0 \) holds.

Put simply, the point is to check at which points the first derivative of the function changes its sign. The extreme values ​​of the function are at these points.

The following procedure is hidden behind the above definitions:

  1. Calculate the first derivative
  2. Calculate zeros of the first derivative
  3. Name intervals
  4. Establish monotony table
  5. Calculate the sign of the intervals
  6. Interpret the result
    \ (\ rightarrow \) if the first derivative changes at a point from a positive to a negative sign, then there is a high point
    \ (\ rightarrow \) if the first derivative changes at a point from a negative to a positive sign, then there is a low point
  7. Calculate the y-coordinates of the high points / low points

example 1

The function \ (f (x) = x ^ 2 \) is to be examined for extreme values.

1.) Calculate the first derivative

\ (f '(x) = 2x \)

2.) Calculate zeros of the first derivative

\ (2x = 0 \ qquad \ rightarrow \ quad x = 0 \)

3.) Name the intervals

The calculated zero divides the relevant area into two intervals.

  1. Interval: \ (\ left] - \ infty; 0 \ right [\)
  2. Interval: \ (\ left] 0; + \ infty \ right [\)

4.) Set up the monotonicity table

The intervals are in the first line of the monotonicity table.

In the second line of the monotonicity table, we note the signs of the intervals in step 5.

The basic structure of the table looks like this:

\ (\ begin {array} {c | cc}
& \ left] - \ infty; 0 \ right [& \ left] 0; + \ infty \ right [\ \ hline
f '(x) & &
\ end {array} \)

5.) Calculate the sign of the intervals

To calculate the sign of an interval, we put any number of the interval into the first derivative.

  • From the interval \ (\ left] - \ infty; 0 \ right [\) we choose the number "-1":
    \ (f '(- 1) = 2 \ cdot (-1) = -2 \ quad \ rightarrow \ text {negative sign} \)

  • From the interval \ (\ left] 0; + \ infty \ right [\) we choose the number "1":
    \ (f '(1) = 2 \ cdot 1 = 2 \ quad \ rightarrow \ text {positive sign} \)

We note these interim results in the monotony table.

\ (\ begin {array} {c | cc}
& \ left] - \ infty; 0 \ right [& \ left] 0; + \ infty \ right [\ \ hline
f '(x) & - & + \
\ end {array} \)

6. Interpret the result

Since at the point \ (x = 0 \) the first derivative of the function changes from a negative to a positive sign, there is a low point there.

7.) Calculate the y-coordinate of the low point

\ (f (0) = 0 ^ 2 = 0 \)

The coordinates of the low point are: (0 | 0).

The function \ (f (x) = x ^ 2 \) is drawn in the coordinate system. In addition, the extreme value (= lowest point) of the function is marked in red.

In the graph it is easy to see how the first derivative of the function changes its sign at the point \ (x = 0 \). Since the graph first falls (negative sign) and then rises (positive sign), it is a low point.

Example 2

The function \ (f (x) = \ frac {2} {3} x ^ 3 + 3x ^ 2 + 4x \) is to be examined for extreme values.

1.) Calculate the first derivative

\ (f '(x) = 2x ^ 2 + 6x + 4 \)

2.) Calculate zeros of the first derivative

The zeros are \ (x_1 = -2 \) and \ (x_2 = -1 \).

3.) Name the intervals

The calculated zeros divide the relevant area into three intervals.

  1. Interval: \ (\ left] - \ infty; -2 \ right [\)
  2. Interval: \ (\ left] -2; -1 \ right [\)
  3. Interval: \ (\ left] -1; + \ infty \ right [\)

4.) Set up the monotonicity table

The intervals are in the first line of the monotonicity table.

In the second line of the monotony table, we note the signs of the intervals in step 5.

The basic structure of the table looks like this:

\ (\ begin {array} {c | ccc}
& \ left] - \ infty; -2 \ right [& \ left] -2; -1 \ right [& \ left] -1; + \ infty \ right [\ \ hline
f '(x) & & &
\ end {array} \)

5.) Calculate the sign of the intervals

To calculate the sign of an interval, we put any number of the interval into the first derivative.

  • From the interval \ (\ left] - \ infty; -2 \ right [\) we choose the number "-3":
    \ (f '(- 3) = 2 \ cdot (-3) ^ 2 + 6 \ cdot (-3) + 4 = 4 \ quad \ rightarrow \ text {positive sign} \)

  • From the interval \ (\ left] -2; -1 \ right [\) we choose the number "-1.5":
    \ (f '(- 1.5) = 2 \ cdot (-1.5) ^ 2 + 6 \ cdot (-1.5) + 4 = -0.5 \ quad \ rightarrow \ text {negative sign} \ )

  • From the interval \ (\ left] -1; + \ infty \ right [\) we choose the number "0":
    \ (f '(0) = 2 \ cdot (0) ^ 2 + 6 \ cdot (0) + 4 = 4 \ quad \ rightarrow \ text {positive sign} \)

We note these interim results in the monotony table.

\ (\ begin {array} {c | ccc}
& \ left] - \ infty; -2 \ right [& \ left] -2; -1 \ right [& \ left] -1; + \ infty \ right [\ \ hline
f '(x) & + & - & +
\ end {array} \)

6. Interpret the result

Since at the point \ (x = -2 \) the first derivative of the function changes from a positive to a negative sign, there is a high point there.

Since at the point \ (x = -1 \) the first derivative of the function changes from a negative to a positive sign, there is a low point there.

7.) Calculate the y-coordinate of the high point / low point

\ (y = f (-2) = \ frac {2} {3} \ cdot (-2) ^ 3 + 3 \ cdot (-2) ^ 2 + 4 \ cdot (-2) = - \ frac {4 } {3} \)

The coordinates of the high point are: \ (\ left (-2 | - \ frac {4} {3} \ right) \).

\ (y = f (-1) = \ frac {2} {3} \ cdot (-1) ^ 3 + 3 \ cdot (-1) ^ 2 + 4 \ cdot (-1) = - \ frac {5 } {3} \)

The coordinates of the lowest point are: \ (\ left (-1 | - \ frac {5} {3} \ right) \).

The function \ (f (x) = \ frac {2} {3} x ^ 3 + 3x ^ 2 + 4x \) is drawn in the coordinate system. In addition, the extreme values ​​of the function are marked in red.

In the graphic you can clearly see how the first derivative of the function changes its sign at the positions \ (x = -2 \) and \ (x = -1 \).

Procedure 1 or Procedure 2?

In this chapter we got to know two methods to calculate the extreme values ​​of a function. The question arises as to which method is best used when.

Reasons for procedure 1 (with second derivative)

If, in addition to calculating extreme values, you are also asked about curvature behavior or turning points in a task, use this procedure. Since you have to calculate the second derivative anyway, you can also use this directly to calculate the extreme values.

Reasons for procedure 2 (without second derivation)

If you do not (!) Need the second derivative in the course of a task, you save yourself having to calculate it and use a monotony table to calculate the extreme values. In the case of fractional rational functions, it can often be very time consuming to write to calculate the second derivative. So it can be worthwhile to do without this if you do not need the second derivative - as I said - in the further course of the task.

Conclusion

Read the assignment completely and consider whether you need the second derivative. This can potentially save you a lot of valuable time.

Extreme values ​​- collection of formulas

In the following overview you will find a collection of formulas for calculating the extreme values.

 condition
Calculate high point\ (f '(x_0) = 0 \ qquad \ text {and} \ qquad f' '(x_0) <0 \)
Calculate low point\ (f '(x_0) = 0 \ qquad \ text {and} \ qquad f' '(x_0)> 0 \)
Calculate the turning point\ (f '' (x_0) = 0 \)
\ (f '' '(x_0) \ neq 0 \)
Calculate saddle point\ (\ left. \ begin {align *} f '' (x_0) & = 0 \ f '' '(x_0) & \ neq 0 \ end {align *} \ right \} \) Condition for a turning point
and
\ (f '(x_0) = 0 \) (condition for a horizontal tangent)

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