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## PowerPoint Slideshow about ' CONTINUITY AND DIFFERENTIABILITY' - ifeoma-waters

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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

Y’

CONTINUITYSuppose 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.

DefinitionQ1) 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.

QUESTIONCase2 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

continuedSuppose 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]

DIFFERERENTIABILITYIf 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

TheoremProvided 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

Derivatives of implicit functionsQ1 Find the derivative of the function xy+y.²=tanx+y

Solu:-

Differentiating both sides w.r.t.x

1.y+xdy/dx+2ydy/dx=

(x+2y-1) =

questionConsider 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

Logarithmic differentiationQ1 DIFFERENTIAT THE FUNCTION W.R.T. X

- Solu:-Let u= and v= taking log on both sides logu=xloglogx and logv=logxlogx
- Differentiating both sides w.r.t.x
- =1.loglogx+x.1/xlogx =2logx/x
- =u(loglogx+1/logx) =v(2logx/x)
- Let y= , y=v+u,

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

Derivatives of functions in parametric formsQ1 If x and y are connected parametrically by equations , without eliminating the parameter , find dy/dx Of x=a(

- Solu:- x=
- = =

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