Question 8

The answer given behind is options a,b,c

I had one doubt that is the function differentiable and continuous at -1 and 3

Also in options a and b square brackets are used. so itmeans only at specific values.

No the function is continuous at x = 2 as left hand limit is equal to right hand limit but the function is not differentiable at x = 2 as left hand derivative is not equal to right hand derivative.

Can anybody answer my doubt:

I want to know is the function continuous and differentiable at x = 1 and x = -3 i.e the ends.

@ oneyeartogo,

 perfectly said, the function is non differentiable at x = 2.

Tangent at a point on the curve is defined as the limiting value (as the the two points where the secant cuts the curve approach each other) of the secant to the curve, but since we do not have the function defined beyond 3, we cant draw a secant hence we cant draw a tangent. Is this right?

"Yeah..it is as we dont have a right & left side respectively at x=1, x = -3, but we definetly have just one tanget at these 2 points as compared to x =2."

What do you mean by "we dont have a right & left side respectively at x=1, x = -3, but we definetly have just one tanget at these 2 points as compared to x =2".

We can have only one tangent at a point. At x = 2 we can have only 1 tangent.

For a function to be differentiable, there needs to be a unique tanget.....which is what is meant by LHD = RHD ( when the slope is same when u approach the point at which u calculate the differential )

If u see any function which is non-differentiable....where  LHD is not equal to RHD, it  means non - existence of a unique tangent...(There will be  a sudden turn in the Function if you plot it graphically...)

like f(x) = |x| ,

LHD = -1

RHD = 1

if u see these are the slopes of the line  -x & x

Same is the case for this 1993 question

How is it continuous at -1 and 3?

Somebody please help.

Let me clear all your doubts now,

1)Every differentiable function is continuous for sure and the function which is continuous may be differentiable or may not be differentiable.

2)Since the function has domain from [-1,3] there is no need to calculate LHL at -1 and RHL at 3 as these points will not lie in domain.

3)Since 3x²+12x-1 is an algebric function it is continuous at x=-1 and continuous also for  -1x2 ,and since 37-x is also an algebric function it is also continuous for x=3 and for 2<x3.

4)Now the only point left to check conituity is 2,

LHL

(lim x tends 2)3x²+12x-1

(lim h tends 0)3(2-h)²+12(2-h)-1

(lim h tends 0)3(4+h²-4h)+24-12h-1=36-1=35

RHL

(lim h tends 0)37-(2+h)=35-h=35

f(2)

3(2)²+12(2)-1=12+24-1=35

since LHL=RHL=f(2) function is continuous at x=2 and for -1x3

@ oneyeartogo,

Hey,,,sorry for late response..

When we talk of continuity on closed interval [X,Y], we need to check on the Left hand side continuity at X and Right hand continuity of Y as correctly mentioned by the dumbheadwithnobrain

Another way to look at is the graphical Approach. Plot the Graph. Then if u are able to reach the end of the domain from the start without lifting your pencil, it implies that f(x) is continious on the Domain which is the case with this 93 question too...

 

Here is a link which you might find useful.

http://www.math10.com/en/algebra/functions/function-continuity/function-continuity.html

Cheers,

Sudhir.

Thanks everybody for replying but I'm still not convinced.

@thedumbheadwithnobrain

2)But -1& 3 lie in the domain of the function as [ ] signifies that the end points are included

3) How can you say it is continuous at -1 and 3 since it is an algebraic function. (I know algebraic functions are continuous everywhere but that is precisly my doubt). From the definition of continuity it has to be discontinuous at x = -1 and x = 3. (explained in detail below)

I'll clearly frame my doubt.

I know we must have LHL = RHL = value of function at that point for a function to be continuous at that point.

But lets say a function y = x is defined for an interval [1,10] .

How can we say it is continuous at x = 1 and x =10?

Left hand Limit for x = 1 is not defined and for x = 10 RHL is not defined. so from the definition of continuity we can say the function is not continuous at x = 1 and x = 2.

Similarly in Question 8 given above we can say function is not continuous at x = -1 and x = 3.