Maimum and Minimum Values y Maimum Minimum MATH 80 Lecture 4 of 6
Definitions: A function f has an absolute maimum at c if f ( c) f ( ) for all in D, where D is the domain of f. The number f (c) is called the maimum value of f on D. (Also called the global maimum.) A function f has an absolute minimum at c if f ( c) f ( ) for all in D. The number f (c) is called the minimum value of f on D. (Also called the global minimum.) y 7 6 5 4 0 0 4 5 6 7 8 9 0 Absolute (Global) Maimum Absolute (Global) Minimum MATH 80 Lecture 4 of 6
A function f has an relative maimum at c if f ( c) f ( ) for all near c. The number f (c) is called the maimum value of f on D. (Also called the local maimum.) A function f has an relative minimum at c if f ( c) f ( ) for all near c. The number f (c) is called the minimum value of f on D. (Also called the local minimum.) The blue dots below represent relative ma and mins. y 7 6 5 4 0 Absolute (Global) Maimum Relative (Local) Minimums Relative (Local) Maimums 0 4 5 6 7 8 9 0 Absolute (Global) Minimum MATH 80 Lecture 4 of 6
Derivatives and the Shapes of Graphs f ( ) = + f ( ) = + f ( ) = + y y y A 5 4 5 4 5 4 B -5-4 - - - 4 5 - A C -5-4 - - - 4 5 - B -5-4 - - - 4 5 - - - C B - - - - -4-4 -4-5 -5-5 MATH 80 Lecture 4 4 of 6
What f () says about f (). The sign of f (). Determines where the function f () increases or decreases. Notice that the function rises to the left of A, falls between A and C, then rises again to the right of C. This information is given in the graph of f (). Notice that the function f () is positive to the left of A, and the right of C. Between A, and C the function f () is negative. Stationary, Maimum, and Minimum Points: The points A and C where the function f () is relatively high and low occur where f ( ) = 0, A, and C. These are called stationary points. Concavity and Inflection: At the point B, the function s concavity changes from concave down to concave up. This corresponds to the point where the graph of f () changes from decreasing to increasing. MATH 80 Lecture 4 5 of 6
Definition A critical number (point) of a function f is a number c in the domain of f such that either f ( c) = 0 or f (c) does not eist. Eample: Find all critical numbers for the function f ( ) =. First f ( ) = = 0, which happens at =. Eample: Find all critical numbers for the function f ( ) =. ( ) () Compute f ( ) = =. ( ) ( ) Note that f () is undefined when =. ( ) Rewrite f ( ) =. So critical numbers are = 0,,. ( ) MATH 80 Lecture 4 6 of 6
Eample: Find all critical numbers for the function f ( ) = + 4. First f ( ) = + 6 4 = 0, so we need to solve + 8 = 0. Factoring is easiest so rewrite this as ( + 4)( ) = 0. Critical points are =, and = 4. Eample: Find all critical numbers for the function f ( ) 4 =. First f ( ) = 4 4, which factors as 4 4 = 4( ) = 4( + )( ). Setting equal to 0, critical points are = 0, ±. Fermat s Theorem If f () has a local maimum or minimum value at c, and if f (c) eists, then f ( c) = 0 MATH 80 Lecture 4 7 of 6
Increasing and Decreasing Test. If f ( ) > 0 on an interval, then f () increases on that interval.. If f ( ) < 0 on an interval, then f () decreases on that interval. First Derivative Test Suppose that c is a critical number of a continuous function f. If f ( ) > 0 for < c, and f ( ) < 0 for > c, then c is a local maimum point. i.e. If f changes from positive to negative at c. If f ( ) < 0 for < c, and f ( ) > 0 for > c, then c is a local minimum point. i.e. If f changes from negative to positive at c. If f does not change sign at c, then f has no local etreme value at c. MATH 80 Lecture 4 8 of 6
MATH 80 Lecture 4 9 of 6
Concavity, and Inflection Points The graph of f () is concave up at = c if the slope function f () is increasing at = c. The graph of f () is concave down at = c if the slope function f () is decreasing at = c. A point where the function changes concavity is called an inflection point. Concavity Test If f ( ) > 0 for all in I, then the graph of f is concave upward on I. If f ( ) < 0 for all in I, then the graph of f is concave downward on I. MATH 80 Lecture 4 0 of 6
5 5 0 Eample: Suppose y ( ) = then y ( ) = and y ( ) = 9 If > 0 then > 0 and y ( ) > 0 so y is concave up there. If < 0 then < 0 and y ( ) < 0 so y is concave down there. Since concavity changes at = 0, (0, 0) is an inflection point. MATH 80 Lecture 4 of 6
5 Eample: Suppose y ( ) = then y ( ) = and y ( ) = 9 If > 0 then y < 0 so y is concave down there. If < 0 then y ( ) > 0 so y is concave up there. MATH 80 Lecture 4 of 6
y ( ) = 4 y ( ) = 4 = ( ) y ( ) = 0 at = 0, y ( 0) = 0 y ( ) = 6 MATH 80 Lecture 4 of 6
Second Derivative Test: Suppose that f is continuous near c. If f ( c) = 0 and f ( c) > 0, then f () has a local minimum at = c. If f ( c) = 0 and f ( c) < 0, then f () has a local maimum at = c. Note: If f ( c) = 0, then anything can happen. f () may have a local minimum, a local minimum, or neither at = c. MATH 80 Lecture 4 4 of 6
Eample: Determine intervals on which the function f ( ) = 4 is increasing or decreasing. Find any local maimum and minimum values, and then intervals on which the function is concave up or concave down and any inflection points. Solution: First, compute the first derivative and set equal to 0: f ( ) = 4 so f = 0 at = 0. So = 0 is a critical number. + + + + + + + + + + + + + + + + - - 0 Sign of f () f < 0 for (, 0) and the function decreases there. f > 0 for ( 0, ) so that is where the function increases. From the diagram there is a relative minimum value of f ( 0) = 4at = 0. Since f = 4, the graph is always concave up, and there are no inflection points. MATH 80 Lecture 4 5 of 6
Steps to Analyzing Graphs of Functions ) Determine critical points i.e. points where f ( ) = 0, or doesn t eist. ) Draw a number line and plot the critical points. ) The critical points divide up the number line into a set of sub intervals. Determine the sign of f () in each of the intervals by checking test points in those intervals. For eample, if the sub interval is [0, ], then a good test point is =. If the sub interval is [, 5] then a good test point is = 0. To determine the sign of f (), plug any test point into f (). Draw + s or s over each interval indicating the sign of f (). 4) Write down the intervals where f () is increasing ( f ( ) > 0) and decreasing ( f ( ) < 0) 5) Check the critical point diagram to determine any relative ma s and/or min s. Mountain tops are relative maimums and valleys are relative minimums. Whenever you want ma s or mins plug your critical points into the original function f ()! MATH 80 Lecture 4 6 of 6
6) Now determine where f ( ) = 0 or doesn t eist. 7) Draw a number line and plot the points in 6). 8) The zeros of f ( ) divide up the number line into a set of sub intervals. Determine the sign of f ( ) in each of the intervals by checking test points in those intervals. For eample, if the sub interval is [, ], then a good test point is = 0. If the sub interval is [, ] then a good test point is =. To determine the sign of f ( ), plug test points into f ( ). Draw + s or s over each interval indicating the sign of f ( ). 9) Write down the intervals where f () is concave up f ( ) > 0 and/or down f ( ) < 0. These are determined by your diagram. 0) Check the f ( ) diagram to determine any points of inflection. A point of inflection is where the concavity changes. To find the actual point plug the - value into f (). ) You now have enough information to graph the function. MATH 80 Lecture 4 7 of 6
Eample: Suppose f ( ) = ) Determine critical points by setting f ( ) = 0 f ( ) = 6 6 so f ( ) = 0 at 6 6 = 6( ) = 0 which gives = 0,. ) Draw a number line and plot the critical points. - - 0 ) Determine the sign of f () in each of the sub-intervals by checking test points in those intervals. For eample; in the sub interval (,0) a good test point is =. Plugging in = gives f ( ) = > 0 so f () is increasing there. Draw + signs above the number line to reflect this. MATH 80 Lecture 4 8 of 6
For the sub interval (0, ) pick =, f ( / ) = < 0. so f () is decreasing there. Draw minus signs above the number line, and for the last interval (, ), pick =. f ( ) = > 0 and so f () is increasing there. The diagram then looks like. (Draw red lines to indicate increases and decreases.) + + + + + + + + + + + + + + + + + + + + - - 0 Sign of f () 4) Write down the intervals where f () is increasing and decreasing. Look at + s and s in the diagram. f () is increasing on the intervals (,0) and (, ). f () is decreasing on the interval (0, ) MATH 80 Lecture 4 9 of 6
5) Check the critical point diagram to determine any relative ma s and/or min s. Since there is a mountain top at = 0, there is a relative ma there with a value of f ( 0) = 0. At = there is a valley, hence there is a relative minimum value of f ( ) =. Whenever you want ma s or mins plug your critical points into the original function! 6) Now determine where f ( ) = 0, which happens when f ( ) = 6 = 0. i.e. at =. 7) Mark this off on a number line. - - 0 8) The zeros of f ( ) divide up the number line into a set of sub intervals. Determine the sign of f ( ) in each of the intervals by checking test points in those intervals. For the sub interval (, ) a good test point is = 0. For sub interval (, ) a good test point is =. To determine the sign of f ( ), plug test points into f ( ). f ( 0) = 6 < 0 and f ( ) = 6 > 0. MATH 80 Lecture 4 0 of 6
Draw + s or s over each interval indicating the sign of f ( ). + + + + + + + + + - - 0 Sign of f ( ) 9) Write down the intervals where f () is concave up and/or down. These are determined by your diagram. f () is concave up on the interval (, ), and concave down on, ) ( 0) Check the f ( ) diagram to determine any points of inflection. A point of inflection is where the concavity changes. To find the actual point plug the - value into f (). Since concavity changes at ( ; ) =, there is an inflection point at ( ; f ( )) or MATH 80 Lecture 4 of 6
) You now have enough information to graph the function. y 5 4 (( 0,0) ( ; ) ( 0 -.0 -.0 0.0.0.0 - - - (( ) -4-5 MATH 80 Lecture 4 of 6
4 Eample: Let f ( ) = 8 + 40. Find intervals on which the function is increasing and decreasing; concave up and down. Locate and classify all etrema, as well as points of inflection. Using this information graph the function. Following the previous steps: Here f ( ) = 4 6, and f ( ) = 6. We can factor the first derivative as f ( ) = 4( + )( ) which has zeros at = 0, ±. and diagram its sign below. (You do this by checking test points 4,,, 4.) + + + + + + + + + + + + -4 - - - 0 4 Sign of f () f () is increasing on the intervals (,0) and (, ) and decreasing on (, ) and ( 0,). =,0, are stationary points. From the diagram f ( ± ) = 4 are relative minimum values and f ( 0) = 40 is a relative maimum value. MATH 80 Lecture 4 of 6
With ( ) f = 6, the zeros are 6 = ( ) = 0, at = ±. Diagramming the second derivative: + + + + + + + + + + + + + + + + -4 - - - 0 4 Sign of f ( ) Concave up on the intervals (, ) and (, ) Concave down on the interval (, ). Since the concavity changes at = ±, there are inflection points there. Note; the actual point of inflection is ( ±,). MATH 80 Lecture 4 4 of 6
Putting it all together: + + + + + + + + + + + + -4 - - - 0 4 Sign of f () + + + + + + + + + + + + + + + + -4 - - - 0 4 Sign of f ( ) MATH 80 Lecture 4 5 of 6
y 50 00 50 00 50 6 5 4 0 4 5 6 50 MATH 80 Lecture 4 6 of 6