officil website http://uoft.me/mat137 MAT137 Clculus! Lecture 20 Tody: 4.6 Concvity 4.7 Asypmtotes Net: 4.8 Curve Sketching 4.5 More Optimiztion Problems
MVT Applictions Emple 1 Let f () = 3 27 20. 1 Find the intervls on which f is incresing or decresing. 2 Find the locl etreme vlues.
Results bout etrem nd monotonicity Theorem (First Derivtive Test) Suppose tht c is criticl point for f nd f is continuous t c, nd tht f is differentible t every point in some intervl contining c ecept possibly t c itself. Moving cross this intervl from left to right, 1 If f f chnges from positive to negtive t c, then f hs locl mimum t c. 2 If f chnges from negtive to positive t c, then f hs locl minimum t c. 3 If f is positive to the left nd right of c, or negtive to the left nd right of c, then f hs no locl mimum or minimum t c.
Results bout etrem nd monotonicity Emple 2 Let f () = 1/3 ( 4). 1 Identify the intervls on which f is incresing or decresing. 2 Find the locl nd globl etreme vlues.
Concvity y y B B A A b b
Concvity y y B B A A b b Definition (Concvity) Let f be function differentible on n open intervl I. The grph of f is 1 concve up on I if f increses on I ; 2 concve down on I if f decreses on I.
Wht does f tell us bout f? Theorem Suppose tht f is twice differentible on n open intervl I. 1 If f () > 0 for ll I, then f increses on I, nd the grph of f is concve up. 2 If f () < 0 for ll I, then f decreses on I, nd the grph of f is concve down.
Concvity Definition (Inflection Point) The point P = (c, f (c)) is clled n inflection point IF f is continuous t c, nd the grph chnges from concve up to concve down or from concve down to concve up t P.
Inflection Point Theorem If the point (c, f (c)) is point of inflection, then f (c) = 0 or f (c) does not eist. WARNING The converse is not true. The curve y = 4 hs no inflection point t = 0. Even though the second derivtive y = 12 2 is zero there, it does not chnge sign.
Concvity Emple 3 Consider the function f () = 1/3. Find the inflection points of the grph of f.
Concvity nd Locl Etrem Theorem Second Derivtive Test for Locl Etrem Suppose f is continuous on n open intervl tht contins c. 1 If f (c) = 0 nd f (c) < 0, then f hs locl mimum t = c. 2 If f (c) = 0 nd f (c) > 0, then f hs locl minimum t = c. 3 If f (c) = 0 nd f (c) = 0, then the test fils. The function f my hve locl mimum, locl minimum, or neither.
Verticl Asymptotes Definition The line = is clled verticl symptote of the curve y = f () if t lest one of the following sttements is true. lim f () = f () = lim lim lim f () = f () = lim lim f () = + + f () =
Verticl Asymptotes Emple 4 Find the verticl symptotes of f () = 3 + 6 2 2 8. y 2
Horizontl Asymptotes Definition The line y = L is clled horizontl symptote of the curve y = f () if either lim f () = L or lim f () = L. y y y = L horizontl sympotote horizontl sympotote y = L
Horizontl Asymptotes Emple 5 Find the horizontl symptotes of the grph of the function f () = 2 + sin. Does the grph of f hve verticl symptotes? A curve my cross one of its horizontl symptotes infinitely often.
More bout Asymptotic Behviour Definition Let f nd g be continuous function. We sy tht f behves like g symptoticlly if lim [f () g()] = 0. ± In prticulr, we sy tht the line y = m + b is n oblique symptote or slnt symptote if lim [f () (m + b)] = 0. ± Remrk For rtionl functions, slnt symptotes occur when the degree of the numertor re more thn the degree of the denomintor. In such cse, the eqution of the slnt symptote cn be found by long division.
Slnt Asymptotes Emple 6 Find the slnt symptotes of the grph of f () = 2 + 1 4 = 2 + 1 2-4 -2 2 4-2 -4 = = -
Asymptotes Emple 7 Find the verticl nd horizontl symptotes of the grph of the function f () = e + 1 e 1.