Math RE - Calculus I Functions Page 1 of 10. Topics of Functions used in Calculus

Size: px

Start display at page:

Download "Math RE - Calculus I Functions Page 1 of 10. Topics of Functions used in Calculus"

Cathleen Alisha Stewart
5 years ago
Views:

1 Math 0-03-RE - Calculus I Functions Page of 0 Definition of a function f() : Topics of Functions used in Calculus A function = f() is a relation between variables and such that for ever value onl one value. A point is represented b the coordinates (,). If we have a set of points like (,) ; (,3) ; (3,) ; (4, ) ; these points represent b definition a function where the domain of the function is the set of values {,, 3, 4} and the range of the function is the set of the corresponding values {,, 3}. Most of the time, a function is presented b an equation. The graph of a function = f() ma have intercepts with the aes. When = 0, we get the intercept and when = 0, we get the intercept. Linear function A linear function has the general function = a + b The graph shows 3 linear functions called lines: () = 3 represents a horizontal line (red) () = + (blue line) (3) = 4 (brown line) Domain of the 3 lines is R (all real numbers) Range of () is {3}. Range of () is R. Range of (3) is R. Even if the window of the graph restricts the view of the domain and range, the lines still Line () shows a horizontal line = = 3 with slope m = 0 and intercept (0,3), no intercept. Line () shows = + with slope m =, intercept (, 0) and intercept (0, ) Line (3) shows = 4 with slope m =, intercept (, 0) and intercept (0, 4) () = 3 3 (3) = 4 () =

2 Math 0-03-RE - Calculus I Functions Page of 0 Equation of a Line: The general form of an equation of a line is = m + b To find the equation, need one of net situations: () coordinates of point (, ) on the line and slope m () coordinates of points (, ) and (, ) Eample : Given point (,3) and slope 4, find the equation of the line. The form used is: = m( ) replace and we get: 3 = 4( ) = = 4 5 The slope is m = 4 and b = 5 ( intercept) Eample : Given points (3,) (,), find the equation of the line. To find the slope m = = 3 ( ) = 4 The form used is: = m( ) replace and we get: = 4 ( + ) = = = 4 + = 7 The slope is m = 4 and b = 7 4 ( intercept)

3 Math 0-03-RE - Calculus I Functions Page 3 of 0 Quadratic function A quadratic function has the general function = a + b + c The graph shows 3 quadratic functions called parabolas: () = (red parabola) () = 4 + (blue parabola) (3) = 3 (brown parabola) Domain of the 3 parabolas is R (all real numbers) Range of () is 0. Range of () is. Range of (3) is. of the domain and range, the parabolas still Parabola () shows = with verte at (0, 0) opening upwards intercept (0, 0) and intercept (0, 0). Parabola () shows = 4 + with verte at (, ) opening upwards intercepts (0.6,0) and (3.4,0) ; intercept (0,). () = (3) = 3 () = + 3 Parabola (3) shows = 3 with verte at (, ) opening downwards no intercept and intercept (0, 3) Eample: Given the curve of = 3, find the verte and the intercepts. Use the quadratic function = a + b + c with a verte at = b a intercept: make = 0 and find the value(s) and find the corresponding value; intercept: make = 0 and find the value In this eample, verte is located at = ( 3) =, = () 3() = so the verte is at (,), opening downwards since a = 3 < 0, intercepts at (0, 0) and (4, 0) ; intercept at (0, 0)

4 Math 0-03-RE - Calculus I Functions Page 4 of 0 Absolute Value function A absolute value function has the general function = a h + k The graph shows 3 absolute value functions called V graphs: () = (red V graph) () = (blue V graph) (3) = + (brown V graph) Domain of the 3 V graphs is R (all real numbers) Range of () is 0. Range of () is. Range of (3) is 0. of the domain and range, the V graphs still V graph () shows = with verte at (0, 0) opening upwards intercept (0, 0) and intercept (0, 0). V graph () shows = with verte at (, ) opening upwards intercepts (,0) and (3,0) ; intercept (0, ). () = () = (3) = + V graph (3) shows = + with verte at (, 0) opening downwards intercept (, 0) and intercept (0, ) Square Root function A square root function has the general function = a h + k The graph shows 3 square root functions called irrational functions: () = (red curve) () = + (blue curve) (3) = + (brown curve) () = + Domain of () is 0 ; Range is 0. Domain of () is ; Range is. Domain of (3) is ; Range is 0. of the domain and range, the irrational curves still curve () shows = has intercept (0, 0) and intercept (0, 0). curve () shows = + has no intercept and intercept (0, ). curve (3) shows = + has intercept (, 0) and intercept (0, ) () = (3) = +

5 Math 0-03-RE - Calculus I Functions Page 5 of 0 Rational function A rational function has the general function = P() Q() () The graph shows a rational function called fractional epression: = = where P() and Q() are polnomials = Domain: R {0} 0 Range: R {0} 0 vertical asmptote: denominator= 0 = = 0 horizontal asmptote: when ± = = 0 no and intercepts. of the domain and range, the rational curve still 0 () The graph shows a rational function called fractional epression: = Domain: R {} Range: R {} vertical asmptote: denominator= 0 = = horizontal asmptote: when ± = = intercept: (,0) and intercept: (0,) of the domain and range, the rational curve still 0 =

6 Math 0-03-RE - Calculus I Functions Page 6 of 0 (3) The graph shows a rational function called fractional epression: = + Domain: R { } Range: R {0} 0 = + vertical asmptote: denominator= 0 = = horizontal asmptote: when ± = = 0 no intercept: and intercept: (0, ) of the domain and range, the rational curve still 0 Piecewise function () The graph shows a piecewise function { if f() = if > Domain: R ; Range: = f() The graph on the left of = is a piece of a parabola The graph on the right of = is a piece of a line of the domain and range, the piecewise curve still 0 () The graph shows a piecewise function { + 4 if g() = if > Domain: R ; Range: R intercept (, 0) ; intercept (0, 4) = g() 4 3 The graph on the left of = is a piece of a parabola The graph on the right of = is a piece of a line of the domain and range, the piecewise curve still 0

7 Math 0-03-RE - Calculus I Functions Page 7 of 0 (3) The graph shows a piecewise function if h() = + if < if > Domain: R ; Range: 3 = h() intercepts (,0) and (,0) ; intercept (0,) The graph on the left of = is a piece of a line The graph between = and = is a piece of a parabola The graph on the right of = is a piece of a line 0 3 of the domain and range, the piecewise curve still 3 Operations on functions There are 6 operations on functions that can be done: () addition ; () subtraction ; (3) multiplication ; (4) division ; (5) composition ; (6) inverse (make sure that ever answer is reduced) () add f() = + to g() = 3 (f + g)() = f() + g() = ( + ) + ( 3) = + () subtract f() = 4 5 from g() = + (f g)() = f() g() = ( + ) (4 5) = (3) multipl f() = 3 with g() = + (f.g) () = f().g() = (3 ).( + ) = = = (4) divide f() = 3 b g() = + ( ) f () = f() g g() = 3 + (5) composition f ( g() ) ( ) = (f g) () or g f() = (g f) () Given f() = 4 and g() = + 5, find (f g) (). replace ever in f() with g() as follows: ( ) (f g) () = f g() = ( + 5) 4( + 5) = ( ) 4 0 =

8 Math 0-03-RE - Calculus I Functions Page 8 of 0 Operations on functions ( ) (5) composition f g() ( ) = (f g) () or g f() = (g f) () Given f() = and g() = 3 + 5, find (g f) (). replace ever in g() with f() as follows: ( ) (g f) () = g f() = 3( ) + 5 = (6) the inverse of f() is denoted b f () To find the inverse of f(), interchange and and we get = f() then isolate to get f (). Given f() =, find f (). Let =, interchange and, we get: = = + = = = + = +, therefore the inverse of f() is f () = +. Note: It is possible that a function has an inverse equation but no inverse function like = f() = has the inverse equation = but this equation is not a function. Eample : Given f() = 3 + 7, find: (a) f() replace ever in f() b, we get: f() = () 3() + 7 = 5 (b) f( + h) f() To get f( + h), replace ever in f() with + h, then subtract f(). f( + h) f() = [ ( + h) 3( + h) + 7 ] [ ] = + h + h 3 3h = h + h 3h Eample : Given f() = 3 +, find: f() f(0) f() = () 3 + () = and f(0) = (0) 3 + (0) = = f() f(0) = ( ) = 3 Eample 3: Given f() =, find: f( + h) f() To get f( + h), replace ever in f() with + h, then subtract f(). f( + h) f() = + h = ( + h) + ( + h) = = ( + h) ( + h) ( + h) h ( + h)

9 Math 0-03-RE - Calculus I Functions Page 9 of 0 Intersection of functions Eample : Given f() = and g() = + 6, find the points of intersection Let f() = g() = = = 0 This quadratic can be solved b factoring or b the quadratic formula. The solution b factoring is ( + 3)( ) = 0 = 3 and = The corresponding values are = 3 = 9 and = = 4 Therefore the points of intersection are ( 3,9) and (,4) 9 6 f() = 4 g() = + 6 The graph of both functions shows these points of intersection Eample : Given f() = 3 + and g() = + 3, find the points of intersection Let f() = g() 3 + = = 0 This equation can be solved b factoring ( + )( ) = 0 = ; = 0 and = The corresponding values are = = 5 ; = 0 = and = = 0 3 f() = 3 + Therefore the points of intersection are (, 5) ; (0, ) and (,) The graph of both functions shows these points of intersection. 5 g() = + 3

10 Math 0-03-RE - Calculus I Functions Page 0 of 0 Division of Polnomial functions Given P() Q() where P() and Q() are polnomial functions. To divide, the degree of P() must be greater than or equal to the degree of Q(). If the degree of P() is less than the degree of Q() ; cannot divide. Once the long division is done, we write the result as: P() remainder = result + Q() Q() Eample : We must use the long division as shown:, the degree of the numerator is and the degree of the denominator is. 3 + ) ( ) 5 where result= 3 and remainder= 5 Thus = Eample : 7 + 4, the degree of the numerator is and the degree of the denominator is. We cannot divide.

Math 111 Lecture Notes

Math 111 Lecture Notes A rational function is of the form R() = p() q() where p and q are polnomial functions. The zeros of a rational function are the values of for which p() = 0, as the function s value is zero where the value