1 Section 2.9 The Mean Value Theorem Rolle's Theorem:("What goes up must come down theorem") Suppose that ² f is continuous on the closed interval [a; b] ² f is di®erentiable on the open interval (a; b) ² f(a) = f(b) Then there is "c" in the open interval (a; b) for which f 0(c) = 0 Geometrically Rolle's Theorem means; if the function values are the same at the end points of a closed interval for a di®erentiable and continuous f, then there is a point c at which the tangent line is horizontal. Check out the picture below Out[58]= a c b Rolle's Theorem is an "existential" theorem it just says there is such a "c", it doesn't say what it is. Also it does not claim "c" to be unique. As you might observe from the graphical example below there could be two such "c"'s (like c and c0 in the graph below)in an interval or even more. 2 Out[21]= c c' Caution The requirements of the Rolle's Theorem are necessary for the theorem to be true. Check out the graph below: Out[1]= The dashed orange line in the graph is connecting the points (¡1; f(¡1)) and (1; f(1)). Clearly, f(¡1) = f(1) and f is continuous over the closed interval [¡1; 1] but there is NO point inside the interval (¡1; 1) at which you can drawn a horizontal tangent line to this graph. Why did the Rolle's Theorem fail? Because f is not di®erentiable over the whole interval (¡1; 1), note that there is a "cusp" at the origin hence f fails to be di®erentiable there. 3 Example Find all numbers "c" in the interval (0; 2) that satisfy the con- clusion to Rolle's Theorem for f(x) = x3 ¡ 3x2 + 2x + 5 Following the "Caution" above let's ¯rst check whether the function f satis- ¯es all the conditions for us to apply Rolle's or not. a) First f(x) is continuous everywhere because it is a polynomial hence it is continuous in particular over the closed interval [0; 2] b) Also f(x) is di®erentiable everywhere because of the same reason that it is a polynomial, hence over the open interval (0; 2). c) f(0) = 0¡0¡0+5 = 5 and f(2) = 8¡12+4+5 = 5 implies f(0) = f(2) = 5 So we may apply Rolle's Theorem. And it guarantees that there is (or are) c(c's) in the open interval (0; 2) such that f 0(c) = 0. That is all. We have to ¯nd what they are by solving the equationp q q 2 0 2 ¡(¡6)§ 6 ¡4(3)(2) 2 1 f (c) = 3c ¡ 6c + 2 = 0 ) c = 2(3) = 1 § 1 ¡ 3 = 1 § 3 You have to make sure that bothq of these c values are in the interval (0q; 2) 1 0 1 which is the case. So for c1 = 1 + 3 , f (c1) = 0 and also for c2 = 1 ¡ 3 , 0 f (c2) = 0 Our main purpose in introducing Rolle's Theorem is to prove the Mean Value Theorem even though Rolle's has other applications. 4 Mean Value Theorem(MVT) If f is continuous on the closed interval [a; b] and di®erentiable on (a; b), then f(b) ¡ f(a) = f 0(c) for some c in (a; b) b ¡ a Here what is going on geometrically: f(b)¡f(a) Note that the quotient b¡a is the slope of the secant line connecting the points (a; f(a)) and (b; f(b)). Mean Value Theorem (MVT) says that there is at least one c (maybe multiple as in the graph above) in the open interval (a; b) where the slope of the tangent line drawn to f is the same as the slope of the secant line. Cautionary remark and example after Rolle's Theorem holds here as well. To be able to apply MVT you have to make sure ¯rst all conditions are satis¯ed. Also, MVT is another existential theorem, meaning it guarantees existence of a number c, it does not tell what it is. Furthermore just as in the graphical example above MVT does not guarantee uniqueness either, we might have couple of such "c"'s in the interval (a; b). You can also interpret MVT physically as follows: If you are traveling from Champaign to Chicago and your average speed during the trip is 60mi=h, then MVT says at some point during your trip you had to be traveling at exactly 60mi=h. 5 Proof of MVT First let's write down the equation of the secant line. f(b)¡f(a) We have observed above that the slope of the secant line is msec = b¡a . Since (a; f(a)) is a point on this line the secant line equation is ysec ¡ f(a) = msec(x ¡ a) or y = msec(x ¡ a) + f(a): Now let h(x) = f(x) ¡ ysec or by the above de¯nition h(x) = f(x) ¡ msec(x ¡ a) ¡ f(a) Next note h(a) = f(a) ¡ msec(a ¡ a) ¡ f(a) = f(a) ¡ 0 ¡ f(a) = 0 and h(b) = f(b) ¡ msec(b ¡ a) ¡ f(b) µ ¶ f(b) ¡ f(a) = f(b) ¡ (b ¡ a) ¡ f(a) b ¡ a = f(b) ¡ (f(b) ¡ f(a)) ¡ f(a) = 0 So h(a) = h(b) = 0. Moreover h(x) is di®erentiable because it is the di®er- ence of two di®erentiable functions.(One of the conditions on f is di®erentia- bility, ysec is a polynomial and hence di®erentiable everywhere.). So h(x) is continuous by a similar reasoning. Therefore, h(x) satis¯es all the following hypothesis of Rolle's Theorem: i) h is continuous on [a; b] ii)h is di®erentiable on (a; b) iii) h(a) = h(b) = 0 Hence by Rolle's Theorem there is a "c" in the open interval (a; b) such that h0(c) = 0. What is h0(x)? 0 0 h (x) = f (x) ¡ msec + 0 At "c" f(b) ¡ f(a) h0(c) = f 0(c) ¡ m = 0 ) m = f 0(c) ) = f 0(c) sec sec b ¡ a 6 Applications of Mean Value Theorem Example Consider the function f(x) = x3 ¡ x on [0; 2]. Show that it satis- ¯es MVT ¯rst and secure the existence of c. Then ¯nd c (if possible). f(x) is a polynomial which is continuous and di®erentiable every where hence in particular on our interval. So by MVT there is a "c" in the interval (0; 2) for which f(2) ¡ f(0) (22 ¡ 2) ¡ 0 = = 3 = f 0(c) 2 ¡ 0 2 ¡ 0 Since f 0(x) = 3x2 ¡ 1 we will be able to ¯nd c exactly in this case: 4 2 f 0(c) = 3 ) 3c2 ¡ 1 = 3 ) 3c2 = 4 ) c2 = ) c = §p 3 3 Be careful here though, only p2 is the only one of these that is in the interval 3 (0; 2). So c = p2 3 Example Prove that f(x) = 2x3 +3x¡1 has exactly on root in the interval [0; 1]. First let's understand the problem well. The questions wants us to prove there is a root in this interval and the root is unique. It does NOT expect us to ¯nd the root. That is a task we will leave for another Chapter. Showing existence and then proving uniqueness will require a united power of two important theorems. We will show existence by using Intermediate Value Theorem and the we will prove the uniqueness of this root by Rolle's Theorem. Existence of the root: Note that f(x) is a polynomial and f(1) > 0 and f(0) < 0, so by Intermediate Theorem there is a root of the polynomial f(x) in the interval (0; 1). Now we know there is at least one root in this interval. But question says there is exactly one. Rolle's Theorem will take care of that part for us as follows; 7 Uniqueness of the root: Suppose f(x) has two (or more) distinct roots in the interval (0; 1). According to the statement of the problem this as- sumption is wrong. So by assuming this if I run into a wall (get into a mathematical trouble I cannot get out), then problems statement is correct. Let's call these two roots as "A" and "B". So f(A) = f(B) = 0. We will form a new interval using A and B [A; B] and apply the Rolle's Theorem on this interval. (We may apply it because f is a polynomial hence it is continuous and di®erentiable everywhere). By Rolle's theorem there is k in the interval (A; B) such that f 0(k) = 0 ) 6k2 + 3 = 0 Note that this equation has no solution, so there is no such k. But Rolle's Theorem promised one. So I hit that "mathematical trouble/wall". I have to go back and accept the defeat and say that there is only one root in the interval [0; 1] as two gave me headache. Example Suppose f(0) = ¡3 and f 0(x) · 5 for all values of x. How large can f(2) possibly be? Assuming f is continuous and di®erentiable apply MVT on some interval that contains 0 and 2. So we get f(2) ¡ f(0) = f 0(c) for some c 2 ¡ 0 f(2) ¡ f(0) = 2f 0(c) f(2) = 2f 0(c) + f(0) Now use the information given on the derivative f 0(x) · 5, so f 0(c) · 5.