Calc1131W7T1 - Counting real zeros with the IVT

Saturday, 21 April 2018

11:24 AM

Machine generated alternative text: The Intermediate Value Theorem (IVT) goes back to the Flemish engineer; inventor; mathematician and land yacht pioneer Simon Stevin Simon Stevin (1548-1620) The 'VT theorem states that iff@) is a continuous function on the interval [a,b] with f (a) < 0 and f (b) > O, then there is some c e [a,b] with f(c) The same conclusion also holds if f(a) > 0 and f (b) < 0. For example if we know that f (x) is a polynomial function and that f(2) — f(4) — —2 and f (5) —3 —1 —1 then we can deduce that f must have at least one real zero somewhere in the intervals (—00, 1) (5, DC) Based on this information alone, the number of real zeros is at least The 'VT tells us that if f is continuous on [a, b] and f(a) and f(b) are have opposite signs (one positive, the other negative), then f has at least one real zero in the interval [a, b] This fact alone is not very useful for counting the exact number of real zeros of f . But, if we also know that the function is strictly increasing (f' (x) > 0) or strictly decreasing ( f'@) < 0 ) on the interval (a, b) then f has To continue the example above, if f'@) > 0 on (1, 3) and f'@) < 0 on (3, 4), then f has exactly 2 O real zeros in the interval [1, 4].

 

Machine generated alternative text: A differentiable function f@) is said to be increasing on the interval I when it is either strictly increasing f' (x) > 0 or constant f'@) = O for all x e I. A increasing function preserves order in the sense that bigger function input results in bigger function output: if a, b e I and a b then f(a) f(b)_ Similarly a function f(x) is said to be decreasing on the interval I when it is either strictly decreasing f' (x) < 0 or constant f'@) = O for all x e L A decreasing function preserves the order in the sense that bigger function input results in smaller function output: if a, b e I and a bthen f(a) 2 f(b) A increasing or decreasing function on I is called monotonic on I. In the special case when I = Domain(f) t we usually replace "on I " with "evemvhere't For example f(x) x IS Increasing a = 28 b = 88 f2(u Toggle the functions fl@), h f3 (x) and f4 (x) using the GeoGebra app above. You can use the sliders to see the points a and b on the x-axis and the corresponding points f(a) and f(b) on the y-axis. Determine which functions are: Increasing everywhere f2(x) Cl f3(x) f4(x) decreasing everywhere f2(x) Cl f3(x) f4(x) monotonic everywhere f2(x) Cl f3(x) f4(x)

 

Machine generated alternative text: Consider the polynomial p@) The derivative of p is 8x + 7 shown below -20 300 200 100 -100 -200 -300 10 20 By factoring this derivative and looking at the points [—10, p(—10)] and [10, p(10)], we see there are three regions where p is monotonic increasing • decreasing increasing Notice that O on the interval [— O on the interval [— O on the interval [ • P(-IO) — -439/3 < o . p(2) = . p(10) = 1081/3 > o By applying the IVT to p on three intervals, we conclude that p(x) has exactly 3 O real roots. Note we need to introduce two extra points [—10, [10, p(10)] because the IVT can not be applied directly to an infinite interval (—00,

 

Machine generated alternative text: How many zeros does f(x) A sensible guess might be 2 cos(aræ) have in the interval I = [14, 16] 14 15 16 O How do you know there are not more? We can use the IVT and monotonic functions to find out We would like to use the IVT on intervals where f@) is monotonic. The function f is not monotonic on I, but we can break I into two subintervals Il U 12 — I where f is monotonic on Il and separately monotonic on 12 Since f'@) • decreasing on the interval Il — • increasing on the interval 12 O , the function is [14, 15] [15,161 We apply the IVT for f on these intervals separately Since f@) is continuous on the interval Il and • f(14) — • f(15) — then by the IVT there is at at least one zero in this interval. Since the function is monotonic on Il there is exactly one zero in the interval Il Similarly, f(x) is continuous on the interval 12 and • f(15) — • f(16) — Then by the 'VT there is at least one zero in 12_ Since the function is monotonic on 12 there is exactly one zero in 12 _ Hence the number of zeros in the interval I expected _ Note the Maple syntax for the interval [l,ff] is [1, pi) Il LJ 12 is exactly as we

 

Machine generated alternative text: Let's find a decimal expansion for the positive root of the continuous function f@) Divide the interval [0, 10] into ten equally sized subintervals. Since —1 < 0 and 2 —2 then by the Intermediate Value Theorem, there is some number 1 < c < 2 such that f(c) = 0 . Furthermore, we know that the first decimal digit of c must be Now subdivide the interval [1, 2] into ten equally sized subintervals (move the zoom slider along to 2). The function f(x) crosses the x axis at the point A which is contained within the inverval (1.4:1 5) 1.4142 ,41422 1.41423 zoom 6 1.41425 1.41424 4 1.41428 1.41429 So the second decimal digit of c is • third decimal digit of cis 1 • fourth decimal digit of cis 4 • fifth decimal digit of cis 2 • sixth decimal digit of cis 1 O Repeating this process we discover that the Historical Note: Simon Stevin (1548-1620) used this method (without GeoGebra) to construct the first decimal approximation of v'ä

 

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