CONTINUITY AND DIFFERENTIABILITY. Consider the function f(x)= 1, if x 0 =2, if x > 0 Graph of this function is Y This function is defined at every
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
CONTINUITY AND DIFFERENTIABILITY
Consider the function f(x)= 1, if x 0
=2, if x > 0
Graph of this function is Y
This function is defined at every
Points of real line.The value of the
Function at nearby points on the (0,2) y=f(x)
X-axis remain close to each other
Except at x=0. At the points near and to the left (0,1)
Of 0 , the value of the function is 1.At X’ X
The points near and to the right of 0 O
The value of the function is 2.L.H.L. of the
Function at 0 is 1 and R.H.L. of the function
At 0 is 2, so L.H.L. IS NOT EQUAL TO R.H.L
Also the value of the function at x=0 is 1=L.H.L. Y
The graph of this function cannot be drawn in one stroke
This function is not continuous at x=0
Suppose f is a real function on a subset of the real number and let c be a point in the domain of f . Then f is continuous at c if f(x)=f(c)
If the left hand limit , right hand limit and the value of the function at x=c exist and equal to each other, then f is said to be continuous at x=c, if f is not continuous at c , then f is discontinuous at c and c is called a point of discontinuity of f.
Q1) Discuss the continuity of the function f defined by f(x)=x+2, if x 1
=x-2, if x>1
Solu:- The function f is defined at all the points of the real line.
Case 1:- If c is less then 1 , then f(c)=c+2
Therefore f(x)= (x+2)=c+2 , thus , f is continuous at all real numbers less than 1.
Case2 If c is greater than 1, then f(c)=c-2
Therefore f(x)= (x-2)=c-2=f(c )
Thus f is continuous at all points greater then 1
Case3 if c=1, then L.H.L. of f at x=1 is y
f(x)= (x+2)=1+2=3 X’ X
R.H.L. of f at x=1 is f(x)= (x-2)=-1
Since L.H.L. Is not equal to R.H.L.of f at x=1 Y’z
So , f is not continuous at x=1
Suppose f is a real function and c is a point in its domain. The derivative of f at c is defined by
[ f(c+h)-f(c)]/h provided this limit exists.Derivative of f at c is denoted by f’(c). The function defined by f’(x)= wherever the limit exists is defined to be the derivative of f . If this limit does not exist the function is not differenciable,i.e. if and are finite and equal, then f is differentiable in the interval[a,b]
If a function f is differentiable at a point c,then it is also continuous at that point.
Every differentiable function is continuous.
CHAIN RULE:-Let f be a real valued function which is a composite of two functions u and v ; i.e. f=vou. Suppose t=u(x) and if both dt/dx and dv/dt exist, then
If f is a real valued function and composite of three functions, u,v,and w , then f=(wou)ov. If t=v(x) and s=u(t), then
Provided all the derivatives in the statement exist.
Consider the function x+sinxy-y=0 , in this y cannot be expressed as a function of x, such type of functions are called implicit functions.
We differentiate these types of function w.r.t.x and then find dy/dx
Q1 Find the derivative of the function xy+y.²=tanx+y
Differentiating both sides w.r.t.x
Consider the function of the form y=f(x)=
By taking logarithm on both sides , rewriting
log y=v(x) log[u(x)]
Using chain rule , we may differentiate this
The relation expressed between two variables x and y in the form x=f(t), y=g(t) is said to be parametric form with t as a parameter.
Derivative of the functions of such form is
provided f’(t) 0