Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 1 of 9

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1 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 1 of 9 Critical numbers - Increasing and decreasing intervals - Relative Etrema Given f(), the derivatives f () and f () give important information about f(). f () will help find the critical numbers (C.N.). To find them, make the answer of f () equal to zero; finall solve for and these value(s) are called C.N. of f(). Use these C.N. for the first derivative test to find the intervals where f() is increasing or decreasing and the relative etrema. Eample 1: Given f() = , find: (a) the critical numbers of f() f () = =12( 1)( 3) f () =0 = 12( 1)( 3) = 0 the critical numbers are =0,=1 and =3 (b) the intervals where f() is increasing or decreasing. Use the first derivative test b locating C.N. on the number line and test f () for each interval. If f () > 0,f() is increasing; f () < 0,f() is decreasing. (c) the relative etrema of f() To find the etrema, eamine each C.N. in the first derivative test to decide: If before the C.N. f () > 0 and after the C.N. f () < 0, then this C.N. is a relative maimum and find the corresponding value. If before the C.N. f () < 0 and after the C.N. f () > 0, then this C.N. is a relative minimum and find the corresponding value. From the first derivative test, we get at = 0, a relative minimum and the value is f(0) = 9 at = 1, a relative maimum and f(1) = 4 and at = 3, a relative minimum with f(3) First Derivative Test Note: f() is a polnomial function without asmptotes. f() is increasing at 0 <<1or>3 f() is decreasing at <0or1<<3 Relative etrema are: relative minimum at (0, 9) and(3, 36) relative maimum at (1, 4)

2 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 2 of 9 Eample 2: Given f() = , find: (a) the critical numbers of f() f () = : numerator= 0 = +4=0 = = 4 ; denominator= 0 = 2 3 =0 = =0 the critical numbers are = 4 and =0 (b) the intervals where f() is increasing or decreasing. Use the first derivative test b locating C.N. on the number line and test f () for each interval. If f () > 0,f() is increasing; f () < 0,f() is decreasing. (c) the relative etrema of f() To find the etrema, eamine each C.N. in the first derivative test to decide: If before the C.N. f () > 0 and after the C.N. f () < 0, then this C.N. is a relative maimum and find the corresponding value. If before the C.N. f () < 0 and after the C.N. f () > 0, then this C.N. is a relative minimum and find the corresponding value. From the first derivative test, we get at = 0, no etrema since the value is undefined, a vertical asmptote at = 0;at = 4, a relative minimum and f( 4) = 1 16 = First Derivative Test Note: f() is a rational function with asmptotes. f() is increasing at 4 <<0 f() is decreasing at < 4 or>0 Relative etrema is: relative ( minimum ) at 4,

3 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 3 of 9 Concavit: Point(s) of inflection Given f(), the derivatives f () and f () give important information about f(). f () will help to find the possible point(s) of inflection (P.I.). To find them, make the answer of f () equal to zero; finall solve for and these value(s) are called possible values P.I. of f(). Use these possible P.I. for the second derivative test to find the intervals of concavit and P.I. Two tpes of concavit (curve shape): concave down concave up f () < 0 f () > 0 Eample 3: Given f() = 3 27, find: (a) the values for possible point(s) of inflection of f() f () = ; f () =6 f () =0 = 6 =0the value for possible P.I. is =0 (b) the intervals where f() is concave up or concave down. Use the second derivative test b locating value(s)of possible P.I. on the number line and test f () for each interval. If f () > 0,f() is concave up; f () < 0,f() is concave down. Second Derivative Test + 0 Note: f() is a polnomial function without asmptotes. (c) an point of inflection for f() To find the point(s) of inflection, eamine each possible P.I. in the second derivative test to decide: If before the possible P.I., f () > 0 and after the possible P.I., f () < 0, then this P.I. is a point of inflection and find the corresponding value. If before the possible P.I., f () < 0 and after the possible P.I., f () > 0, then this P.I. is a point of inflection and find the corresponding value. From the second derivative test, we get at = 0, a point of inflection and the value is f(0) = 0 Point of inflection at (0, 0)

4 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 4 of 9 f() is concave up at >0 f() is concave down at <0 Point of inflection is: (0, 0) 0 Eample 4: Given f() = 32 ( +2) 2, find: (a) the values for possible point(s) of inflection of f() f () = 12 ( +2) 3 ; f 24( 1) () = ( +2) 4 from f () : numerator= 0 = 24( 1) = 0 = = 1 ; denominator= 0 = ( +2) 4 =0 = = 2 the values for possible P.I. are = 2 and =1 (b) the intervals where f() is concave up or concave down. Use the second derivative test b locating value(s)of possible P.I. on the number line and test f () for each interval. If f () > 0,f() is concave up; f () < 0,f() is concave down. (c) an point of inflection for f() To find the point(s) of inflection, eamine each possible P.I. in the second derivative test to decide: If before the possible P.I., f () > 0 and after the possible P.I., f () < 0, then this P.I. is a point of inflection and find the corresponding value. If before the possible P.I., f () < 0 and after the possible P.I., f () > 0, then this P.I. is a point of inflection and find the corresponding value. From the second derivative test, we get at = 2, no point of inflection since the value is undefined, a vertical asmptote at = 2; at = 1, a point of inflection and the value is f(1) = 1 3 Point of inflection at ( ) 1, 1 3 Second Derivative Test Note: f() is a rational function with asmptotes. f() is concave up at < 2 or 2 <<1 f() is concave down at >1 Point of inflection is: ( ) 1,

5 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 5 of 9 Relative Etrema and the second derivative test Given f(), we saw that the first derivative test finds the relative etrema and the second derivatives f () will do the same. We need f () to find the critical numbers (C.N. ). Replace these C.N. in the answer of f (). If f (C.N.) > 0, then C.N. is a relative minimum and find the corresponding value. If f (C.N.) < 0, then C.N. is a relative maimum and find the corresponding value. If f (C.N.) = 0 or undefined, then C.N. is not a relative etremum. Eample 5: Given f() = , find: the relative etrema using the second derivative test. f () = =12( 1)( 3) f () =0 = 12( 1)( 3) = 0 the critical numbers are =0,=1 and =3 f () = replace =0 = f (0) = 36 > 0, relative minimum at (0, 9) replace =1 = f (1) = 24 < 0, relative maimum at (1, 4) replace =3 = f (3) = 72 > 0, relative minimum at (0, 36) Note : These results are the same as in Eample 1 and the graph of Eample 1. Eample 6: Given f() = , find: the relative etrema using the second derivative test. f () = : numerator= 0 = +4=0 = = 4 ; denominator= 0 = 2 3 =0 = =0 the critical numbers are = 4 and =0 f () = +6 4 replace = 4 = f ( 4) = > 0, relative minimum at ( 4, 1 16 replace =0 = f (0) = undefined, no etrema since the value is undefined, a vertical asmptote at = 0 Note : These results are the same as in Eample 2 and the graph of Eample 2. )

6 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 6 of 9 Absolute Etrema Given f() over a closed interval [a, b], find the highest value in that interval called the absolute maimum and the lowest value in that interval called the absolute minimum. First we locate an C.N. in that interval using the first derivative f (), then we build a table of values including the boundaries of the interval and an C.N. in that interval. The table has the values and the corresponding values. The highest will be the absolute maimum and the lowest will be the absolute minimum. Eample 7: Given f() = 3 27 over [0, 5] f () = = 3( 3)( +3) the critical numbers are = 3 ; =3;onl = 3 is between =0 and =5 = f() 0 f(0) = 0 = absolute maimum 3 f(3) = 54 = absolute minimum 5 f(5) = 10 Eample 8: Given f() = 2 on [ 2, 6] ( +1) 2 f () = 5 ( +1) 3 numerator= 0 = 5 =0 = = 5 ; denominator= 0 = ( +1) 3 =0 = = 1 the critical numbers are = 1 ; = 0 ; both values are between = 2 and =6 = f() 2 f( 2) = 4 1 f( 1) = 5 6 f(5) = f(6) = absolute maimum is at = 5 ; no absolute minimum.

7 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 7 of 9 Curve Sketching To sketch the curve of f() using the derivative method implies following these steps: (1) f() gives intercepts, an asmptotes if f() is a rational function. (2) f () gives relative etrema, intervals of increasing & decreasing with the first derivative test. (3) f () gives point(s) of inflection, intervals of concavit with the second derivative test. (4) a table including all results from steps (1) to (3). (5) Sketch the curve of f() showing all points found in steps (1) to (3). Eample 9: Given f() = 2 ; list (if an) all and intercepts, vertical and horizontal asmptotes, relative +2 etrema, points of inflection, intervals where f() is increasing, decreasing, concave up, concave down and then sketch the graph of f(). from f() = intercept: (0, 0) asmptotes: vertical asmptote: denominator= 0 = + 2 = 0 = = 2 2 horizontal asmptote: lim =D.N.E. ; no horizontal asmptote +2 from f ( +4) () = : the critical point at: =0,=0 ( +2) 2 from f 8 () = : no point of inflection ( +2) 3 values: f () + + f () + + values: 8 und 0 shape: inc.down dec.down dec.up inc.up Ma VA Min

8 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 8 of 9 Eample 10: Given f() = ; list (if an) all and intercepts, vertical and horizontal asmptotes, relative etrema, points of inflection, intervals where f() is increasing, decreasing, concave up, concave down and then sketch the graph of f(). from f() =2 2 (9 ) = intercept: (0, 0) ; (9, 0) no asmptotes since f() is a polnomial. from f () = =6(6 ) : the critical points at: (0, 0) ; (6, 216) from f () =36 12 : possible point of inflection at (3, 108) values: f () + + f () + + values: shape: dec.up inc.up inc.down dec.down Min P.I. Ma

9 Math RE - Calculus I Application of the derivative (1) Curve Sketching Page 9 of 9 Eample 11: Sketch f() with the following conditions: points at ( 1, 1), (0, 0), (1, 1) ; horizontal asmptote: lim f() =0 < 1 = f () < 0,f () > 0 1 <<0 = f () < 0,f () > 0 0 <<1 = f () < 0,f () < 0 >1 = f () > 0,f () < 0 intervals signs of signs of shape of f() values f () f () and values ], 1[ + decreasing, concave up 1 ( 1, 1) ] 1, 0[ decreasing, concave down 0 point of inflection at (0, 0) ]0, 1[ decreasing, concave down 1 relative minimum at (1, 1) ]1, + [ + increasing, concave down + horizontal asmptote at =