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Rolle’s Theorem Statement Let a function f:[a,b]R be such that (i) f is continuous on [a,b], (ii) f is differentiable on (a,b), and (iii) f(a)=f(b).Then there exists at least one point c(a,b) such that f )=0. If f crosses the axis twice, somewhere between the twocrossings, the function is flat. The accurate statement of this``obvious observation is Rolles Theorem.
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Interpretation of Rolle’s TheoremGeometrical Interpretation:If the function y=f(x) has a graph which is a continuous curve on[a,b], and the curve has a tangent at every point of (a,b), andf(a)=f(b), then there must exist at least one point c in (a,b) such thatthe tangent to the curve at (c,f(c)) is parallel to the x-axis.
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Algebraic interpretation:If the function f(x) satisfies all the conditions of Rolle’s Theorem,then between two zeros a and b of f(x) there exists at least one zeroof f (x) [i.e., between two roots a and b of f(x) =0 there must exist atleast one root of f (x)=0].Immediate conclusion:Between two consecutive roots of f (x)=0, there lies at most one rootof f(x)=0.Important note:The set of conditions in the Rolle’s Theorem is a set of sufficientconditions.The conditions are by no means necessary.Illustrations: i) The function f defined in [0,1] as follows: f(x)=1, 0 =2, . The function f(x) satisfies none of the conditions of Rolle’s Theorem, yet f(x)=0 for many points in [0,1].
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ii) The function f defined in [-1,3] as follows: f(x)= + . The function f(x) satisfies none of the conditions of Rolle’s Theorem, yet f(x)=0 for many points in [-1,3], In fact f(x)=0 for all x(0,1). The above two examples show that the conditions in the Rolle’s Theorem are not absolute necessity for f(x) to be zero at some point in the concerned interval.iii) The function f defined in [-1,1] as follows: f(x)= . Here, f(x) is continuous on [-1,1], f(-1)=f(1), but f(0) does not exists i.e. , f(x) is not differentiable on (-1,1). Also, f(x) vanishes nowhere in (-1,1)-{0}. Failure of Rolle’s Theorem can be explained by the fact that is not derivable in -1<x<1, though other conditions are satisfied.iv) The function f defined in [-1,1] as follows: f(x)=x3+5x-4. f(x) is continuous on [-1,1], f(x) is derivable on (-1,1), but f(-1)≠f(1). Also, f(x) =3x2+5 vanishes nowhere in (-1,1). Failure of Rolle’s Theorem can be explained by the fact that , though other conditions are satisfied.
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v) The function f defined in [-1,1] as follows: f(x)=x2, x=-1 = 5x , -1<x<1 = x2, x=1 Here, f(x) is continuous on (-1,1) and discontinuous at the points x=-1 and x=1. i.e., f(x) is not continuous on [-1,1], f(x) is derivable on (-1,1), and f(-1)=f(1). f(x) vanishes nowhere in (-1,1). Failure of Rolle’s Theorem can be explained by the fact that f(x) is not continuous in -1 1, though other conditions are satisfied.Examples (iii), (iv)& (v) show that if we drop any of the conditions inRolle’s Theorem then Rolle’s Theorem fails to be true. Important Note: If f(x) satisfies all the conditions of Rolle’s Theorem in [a,b] then the conclusion that f (c)=0 where a<c<b is assured, but if any of the conditions are violated then Rolle’s Theorem will not be necessarily true; it may still be true but the truth is not ensured.
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Lagrange’s Mean Value Theorem (First Mean Value Theorem of Differential Calculus) Statement Let a function f:[a,b]R be such that i) f is continuous on [a,b], ii) f is differentiable on (a,b).Then there exists at least one point c(a,b) such thatAnother Form of Lagrange’s MVT:If in the statement of the theorem, b is replaced by a+h, (h>0), thenthe number between a and b may be written as a+θh, where0<θ<1.Then the theorem takes the following form: Let f:[a,a+h]R be such that iii) f is continuous on [a,a+h], iv) f is differentiable on (a,a+h).Then there exists a real number θ lying between 0 and 1 such that
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or, f(a+h)=f(a)+hf′(a+θh).** Mean Value Theorem relates the mean rate of changeto instantaneous rate of the function.Somewhere inside a chord, the tangent to f will be parallel tothe chord. The accurate statement of this common-senseobservation is the Mean Value Theorem
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For any function that is continuous on [a, b] and differentiableon (a, b) there exists some c in the interval (a, b) such thatthe secant joining the endpoints of the interval [a, b] isparallel to thetangent at c.Geometrical Interpretation:If the function y=f(x) has a graph which is a continuous curve on[a,b], and the curve has a tangent at every point of (a,b),then thereexists at least one point c in (a,b) such that the tangent to the curveat (c,f(c)) is parallel to the line segment joining the points (a,f(a)) and(b,f(b)).Remarks: i) The fraction measures the mean (or average) rate of increase of the function in the interval [a,b] of length b-a. Hence the theorem expresses the fact that, under the conditions stated, the mean rate of increase in any interval [a,b] is equal to the actual rate of increase at
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some point c within the interval (a,b). For instance, the mean velocity of a moving car in any interval of time is equal to the actual velocity at some instant within the interval. This justifies the name Mean Value Theorem. ii) Mean Value Theorem proposes that any differentiable function defined over an interval has a mean value at which a tangent line is parallel to the line joining the end points of the function’s graph on that interval. iii) Rolle’s Theorem is a particular case of Lagrange’s Mean Value Theorem. If f(a)=f(b) holds in addition to the two conditions of Mean Value Theorem then f(b)-f(a)=0 and consequently f (c)=0.Important deductions: i) Let f:[a,b]R be such that *f is continuous on [a,b], *f is differentiable on (a,b) And * f(x)=0 for all x(a,b). Then f(x) is a constant function on [a,b]. ii) Let f:[a,b]R be such that *f is continuous on [a,b], *f is differentiable on (a,b) and * f(x) 0 for all x(a,b). Then f(x) is a monotone increasing function on [a,b].
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iii) Let f:[a,b] R be such that *f is continuous on [a,b], *f is differentiable on (a,b) and * f(x) 0 for all x(a,b). Then f(x) is a monotone decreasing function on [a,b]. Cauchy’s Mean Value Theorem: (Second Mean Value Theorem of Differential Calculus) Statement Let the functions f:[a,b]R and g:[a,b]R be such that i) both f,g are continuous on [a,b], ii) both f,g are differentiable on (a,b), and iii) f(x)≠0 for x(a,b).Then there exists at least one point (a,b) such that .
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Second Statement: Let the functions f:[a,a+h]R and g:[a,a+h]R be such that iv) both f,g are continuous on [a,a+h], v) both f,g are differentiable on (a,a+h), and vi) f(x)≠0 for x(a,a+h).Then there exists at least one real number θ lying between 0 and 1such that .Geometric Interpretation of Cauchy’sMVT:First Interpretation: The functions f and g can considered asdetermining a curve in the plane by means of parametric equationsx=f(t),y=g(t),where atb. Cauchy’s MVT concludes that a point(f(),g())of the curve for some t= in (a,b) such that the slope of theline segment joining the end points (f(a),g(a)) and (f(b),g(b)) of thecurve is equal to the slope of the tangent to the curve at t=.
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Second Interpretation: We may writeHence, the ratio of the mean rates of increase of two functions in aninterval [a,b] is equal to the ratio of the actual rates of increases ofthe functions at some point within the interval (a,b).REMARKS: i) Lagrange’s MVT can be deduced from Cauchy’s MVT by taking f(x)=x, x(a,b). ii) Rolle’s Theorem can be obtained from Cauchy’s MVT by letting f(x)=x and g(b)=g(a). [We have used Rolle’s Theorem to prove Cauchy’s MVT.]
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iii) Both f and g satisfy the conditions of Lagrange’s MVT. Consequently there exist points c and d in (a,b) such that and c and d are different points in (a,b) in general, and therefore a single point in (a,b) may not be found to satisfy the conclusion of Cauchy’s MVT.Some problems: Q1. Show that the equation 4x5+x3+7x-1=0 has exactly one real root. Q2. A function f is differentiable on [0,2] and f(0)=0,f(1)=1,f(2)=1. Prove that f(c)=0 for some c in (0,2). Q3. Prove that the equation (x-1)3+(x-2)3+(x-3)3+(x- 4)3=0 has only one real root. Q4. Prove that between any two real roots of excosx+1=0 there is at least one real root of exsinx+1=0. Q5. Verify Lagrange’s MVT for the function f(x) =x(x-1)(x-2) in the interval [0, ]. Q6. Show that if 0<u<v deduce that < < .
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Q7. If in the Cauchy’s MVT we take f(x)= and g(x)= ,then prove that is the arithmetic mean between a and b.(Refer to the statement of Cauchy’s MVT).Q8. If f(x)= (x-a)m(x-b)n where m and n are positive integers,show that c in Rolle’s theorem divides the segment a x b inthe ratio m:n.Q9. Apply MVT to prove that lies between 10 and10.05.Q10. Show that , where 0< .
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