CONTINUITY
1. CONTINUOUS FUNCTIONS:
If limh→0f(a−h)=limh→0f(a+h)=f(a)=limh→0f(a−h)=limh→0f(a+h)=f(a)= finite and fixed quantity (h>0)(h>0).
i.e. LHL|x=a=RHL|x=a=LHL|x=a=RHL|x=a= value of f(x)|x=a=f(x)|x=a= finite and fixed quantity.
At isolated points functions are considered to be continuous.
2. CONTINUITY OF THE FUNCTION IN AN INTERVAL:
(a) A function is said to be continuous in (a,b)(a,b) if ff is continuous at each & every point belonging to (a,b)(a,b).
(b) A function is said to be continuous in a closed interval [a,b][a,b] if :
∘f∘f is continuous in the open interval (a,b)(a,b).
∘f∘f is right continuous at 'a' i.e. limx→a+f(x)=f(a)=alimx→a+f(x)=f(a)=a finite quantity.
∘f∘f is left continuous at 'b' i.e. limx→h−f(x)=f(b)=alimx→h−f(x)=f(b)=a finite quantity.
Note :
(i) All Polynomials, Trigonometrical functions, exponential & Logarithmic functions are continuous in their domains.
(ii) If ff & g g are two functions that are continuous at x=cx=c then the function defined by : F1(x)=f(x)±g(x);F2(x)=Kf(x),KF1(x)=f(x)±g(x);F2(x)=Kf(x),K any real number, F3(x)=f(x)⋅g(x)F3(x)=f(x)⋅g(x) are also continuous at x=cx=c. Further, if g(c)g(c) is not zero, then F4(x)=f(x)g(x)F4(x)=f(x)g(x) is also continuous at x=cx=c.
(iii) If ff and gg are continuous then fog and gof are also continuous.
(iv) If ff and gg are discontinuous at x=cx=c, then f+g,f−g,f.gf+g,f−g,f.g may still be continuous.
(v) Sum or difference of a continuous and a discontinuous function is always discontinuous.
3. REASONS OF DISCONTINUITY:
(a) Limit does not exist i.e. limx→a−f(x)≠limx→a+f(x)limx→a−f(x)≠limx→a+f(x)
(b) limx→af(x)≠f(a)limx→af(x)≠f(a) Geometrically, the graph of the function will exhibit a break at x=ax=a, if the function is discontinuous at x=ax=a. The graph as shown is discontinuous at x=1,2x=1,2 and 3 .
4. THE INTERMEDIATE VALUE THEOREM :
Suppose f(x)f(x) is continuous on an interval II and aa and bb are any two points of I. Then if y0y0 is a number between f(a)f(a) and f(b)f(b), their exists a number cc between aa and bb such that f(c)=y0.f(c)=y0.
Note that a function ff which is continuous in [a,b][a,b] possesses the following property.
If f(a)f(a) & f(b)f(b) posses opposite signs, then there exists atleast one solution of the equation f(x)=0f(x)=0 in the open interval (a,b).(a,b).
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